1.1 analytical chemistry

2

1.1 Analytical Chemistry

Analytical chemistry deals with methods for the identification of one or more of

the components in a sample of matter and the determination of the relative amounts of

each. The identification process is called a qualitative analysis while the determination

of amount is termed a quantitative analysis. We will deal largely with the latter.

The results of a quantitative analysis are expressed in such relative terms as the

percent of the analyte (the substance being determined) in the sample, the parts of

analyte per thousand, per million, or even per billion parts of sample, the grams or

milliliters of analyte per liter of sample, or the mole fraction of the analyte in the

sample.

Most quantitative analytical measurements are preformed on solutions of the

sample, therefore the study of analytical chemistry makes use of solution concepts

with which the student should have considerable familiarity.

Quantitative analysis is classified into two types of analysis:

1- Volumetric analysis, which concentrates on the exact volume measurement of the

solution during titration. The volumetric methods of analysis include acid-base

titration, precipitation titration, oxidation-reduction titration, and complex formation

titration.

2- Gravimetric analysis, which based upon the measurement of the weight of a

substance of known composition that is chemically related to the analyte. There are

two types of gravimetric analysis, precipitation methods and volatilization method.

1.2 The Chemical Composition of Solutions

Both aqueous and organic solvents find widespread use in chemical analysis.

Nonpolar solvents, such as hydrocarbons and halogenated hydrocarbons, are employed

when the analyte itself is nonpolar. Organic solvents, such as alcohols, ketones, and

ethers, which are intermediate in polarity and which form hydrogen bonds with

solutes, are considerably more useful than their less polar counterparts because they

dissolve a larger variety of both organic and inorganic species. Aqueous solvents,

3

including solutions of the common inorganic acids and bases, are perhaps the most

widely used of all for analytical purposes. Our discussion will therefore focus on the

behavior of solutes in water: reactions in nonaqueous polar media will be considered

in less detail.

Solutions are classified according to the nature of particles of the solute to: true

solution, suspended solution, and colloidal solution.

1- True solution, in which the solute disappears between the molecules of the solvent,

like NaCl in water.

2- Suspended solution, in which the particles of the solute can be distinguished. The

solute particles are separated and settled in the bottom of the container, and do not

pass through filter paper.

3- Colloidal solution, in which the particles of solute are suspended but do not settle

in the bottom of the container, and pass through filter paper.

The presence of the solutes affects the properties of the solvent. They lower the

vapour pressure, therefore the temperature increases above its boiling point to reach its

usual vapour pressure. The boiling points of the solvent also increase in the presence

of the solute, and if the solute is ionic its effect will doubled. Pure water boils at

100oC, but in the presence of the solute it boils at higher temperature. The presence of

the solute also lowers the freezing point of the solvent. Pure water freezes at 00C,

while the presence of sugar for example it freezes at -1.86oC.

1.3 Electrolytes and Non-electrolytes

Electrolytes are solutes which ionize in a solvent to produce an electrically

conducting media. Strong electrolytes ionize completely whereas weak electrolytes are

only partially ionized in the solvents (Table 1.1).

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Table 1.1 Classification of electrolytes

Strong electrolytes Weak electrolytes

1- Inorganic acids HNO3, HClO4,

H2SO4, HCl, HI, HBr, HClO3, HBrO3

1- Many inorganic acids such as

H2CO3, H3PO4, H2S, H2SO3, H3BO3

2- Alkali and alkaline-earth hydroxides

2- Most organic acids

3- Most salts 3- Ammonia and most organic bases

4- Halides, cyanides, and thiocyanates

of Hg, Zn, and Cd

Non-electrolytes are solutes which do not ionize in their solvents, and therefore the

solution does not conduct electricity. Examples are solutions of sugar, and alcohol in

water.

1.4 Self-ionization of Solvents

Many common solvents are weak electrolytes which react with themselves to form

ions (this process is termed autoprotolysis). Some examples are:

2H2O H3O+ + OH-

2CH3OH CH3OH2+ + CH3O

-

2HCOOH HCOOH2+ + HCOO-

2NH3 NH4+ + NH2

-

The positive ion formed by the autoprotolysis of water is called the hydronium ion,

the proton being bonded to the parent molecule via a covalent bond involving one of

the unshared electron pairs of the oxygen. Chemists use the term H+ instead of H3O+.

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1.5 Acids and Bases

The classification of substances as acids or bases was founded upon several

characteristic properties that these compounds impart to an aqueous solution. Typical

properties include the red and blue colors that are associated with the reaction of acids

and bases with litmus, the sharp taste of a dilute acid solution, the bitter taste and

slippery feel of a basic solution, and the formation of a salt by interactions of an acid

with a base.

*Arrhenius acids and bases

Arrhenius defined acids as hydrogen-containing substances that dissociate into

hydrogen ions and anions when dissolved in water:

HCl H+ + Cl-

CH3COOH H+ + CH3COO-

and bases as compounds containing hydroxyl groups that give hydroxides ions and

cations upon the same treatment:

NaOH Na+ + OH-

Al(OH)3 Al3+ + 3OH-

The relative strengths of acids and bases could be compared by measuring the

degree of dissociation in aqueous solution. A completely ionized acid called strong

acid, like HCl, and a partially ionized acid called weak acid, like CH3COOH. The

same rule applied for strong and weak bases.

* Bronsted and Lowry acids and bases

Bronsted and Lowry proposed independently in 1932 that an acid is any substance

that is capable of donating a proton: a base is any substance that can accept a proton.

The loss of a proton by an acid gives rise to an entity that is a potential proton acceptor

and thus a base; it is called the conjugated base of the parent acid. The reaction

between an acid and water is a typical example:

6

H2O + acid conjugate base + H3O+

It is important to recognize that the acidic character of a substance will be observed

only in the presence of a proton acceptor; similarly basic behavior requires the

presence of a proton donor. Neutralization in the Bronsted-Lowry sense can be

expressed as:

acid 1 + base 2 base 1 + acid 2

This process will be spontaneous in the direction that favors production of the weaker

acid and base. The dissolving of many solutes can be regarded as neutralizations, with

the solvent acting as either a proton donor or acceptor. Thus:

acid 1 + base 2 base 1 + acid 2

HCl + H2O Cl- + H3O+

CH3COOH + H2O CH3COO- + H3O+

NH4+ + H2O NH3 + H3O

+

H2O + NH3 OH- + NH4+

Note that acids can be anionic, cationic, or electrically neutral. It is also seen that

water acts as a proton acceptor (a base) with respect to the first three solutes and as a

proton donor or acid with respect to the last one; solvents that possess both acidic and

basic properties are called amphiprotic.

Acids and bases differ in the extent to which they react with solvents. The reaction

between hydrochloric acid and water is essentially complete; this solute is thus classed

as acid in the solvent water. Acetic acid and ammonium ion react with water to a

lesser degree, with the result that these substance are progressively weaker acids.

The extent of reaction between a solute acid or base and a solvent is also

dependent upon the tendency of the latter to donate or accept protons. Thus, for

example perchloric, hydrochloric and hydrobromic acids are all classed as strong acid

7

in water. If glacial acetic acid, a poorer proton acceptor is used as the solvent instead,

only perchloric acid undergoes complete dissociation and remains a strong acid, the

process can be expressed by the equation:

HClO4 + CH3COOH ClO4- + CH3COOH2

+

Acid 1 base 2 base 1 acid 2

Because they undergo only partial dissociation, hydrochloric acid and hydrobromic

acids are weak acids in glacial acetic acid.

A consequence of the Bronsted theory is that the most effective proton donors (that

is the strongest acids) give rise, upon loss of their protons to the least effective proton

acceptors (the weakest conjugate bases).

The general solvent theory includes not only species that qualify as acids or bases

in the Bronsted-Lowry sense but also extends the concept of acid-base behavior to

solvents that do not necessarily contain protons.

* Lewis acids and bases

Lewis defined an acid as an electron-pair acceptor and a base as an electron-pair

donor.

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SO3 + O2- SO42-

base acid

SO3 + OH- HSO4-

base acid

-4AlCl -+ Cl 3AlCl

acid base

* Salts

Salts are formed by the reactions of cations and anions. Some of the salts are

anhydrous like NaCl, KCl, KMnO4 and K2Cr2O7. Other salts are hydrous such as

CaCl2.2H2O, CuSO4.5H2O and ZnSO4.7H2O. Salts exist in its solid state as ions,

therefore, sodium chloride is ionized in its crystalline case into Na+ which is

surrounded by six ions of Cl-, and each Cl- is surrounded by six ions of Na+. These

ions are attached to each other by electrostatic strengths. Thus, these salts are

completely ionized in solvents of dielectric constant like water.

1.6 Chemical Units of Weight

In the laboratory, the mass of a substance is ordinarily determined in such metric

units as the kilogram (kg), the gram (g), the milligram (mg), the microgram (µg), the

nanogram (ng), or the picogram (pg).

g = 103 mg = 106 µg = 109 ng = 1012 pg = 10-3 kg

For chemical calculations, however, it is more convenient to employ mass units

that express the weight relationship or stoichiometry among reacting species in terms

of small whole numbers. The gram formula weight, the gram molecular weight, and

the gram equivalent weight are employed in analytical work for this reason. These

terms are often shortened to the formula weight, the molecular weight and the

equivalent weight.

9

One molecular weight of a species contains 6.02×1023 particles of that species; this

quantity is frequently referred to as the mole. In a similar way, the formula weight

represents 6.02×1023 units of the substance, whether real or not, represented by the

chemical formula.

Example 1

A 25.0 g sample of H2 contains:

25.0 g

= 12.4 moles of H2

2.016 g/mole

6.02×1023 molecules

12.4 moles × = 7.47×1024 molecules H2

Mole

The same weight of NaCl contains:

25.0 g

= o.428 fw NaCl

58.44 g/fw

which corresponds to 0.428 mole Na+ and 0.428 mole Cl-

1.7 Equivalent Weight

It is the mass of a given substance which will combine with or displace a fixed

quantity of another substance.

For acids:

It is the weight of acid that contains one equivalent of a proton.

molar mass

equivalent weight = ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ

no. of equivalent hydrogen ions

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Example 2

Calculate the equivalent weights for the following acids: HCl, H2SO4, H3PO4. Atomic

weights for H = 1, O = 16, Cl = 35.5, S = 32, P = 31.

molar mass

equivalent weight for HCl = ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ

no. of equivalent hydrogen ions

1 + 35.5

ــــــــــــــــــــ = = 36.5 gram/equivalent

1

(2×1) + 32 + (16×4)

for H2SO4 = 49 = ــــــــــــــــــــــــــــــــــــــــــ

2

(3×1) + 31 + (16×4)

for H3PO4 = 32.67 = ــــــــــــــــــــــــــــــــــــــــــ

3

For bases:

It is the weight of base that contains one equivalent of an hydroxide.

molar mass

equivalent weight = ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ

no. of equivalent hydroxide ions

Example 3

Calculate the equivalent weights for the following bases: NaOH, Ca(OH)2, Al(OH)3.

Atomic weights for H = 1, O = 16, Na = 23, Ca = 40, Al = 26.98.

molar mass

equivalent weight for NaOH = ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ

no. of equivalent hydroxide ions

23 + 16 + 1

ـــــــــــــــــــــــــــــــ = = 40 gram/equivalent

1

11

40 + [(16 + 1)]2

for Ca(OH)2 = 37 = ـــــــــــــــــــــــــــــــــــ

2

26.98 + [(16 + 1)]3

for Al(OH)3 = 25.99 = ـــــــــــــــــــــــــــــــــــــ

3

For salts:

It is the weight of the salt that contains the equivalent weight of one of its ions.

molar mass

equivalent weight for salt = ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ

no. of metal ions × oxidation no.

molar mass

ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ =

no. of acidic radical ions × oxidation no.

Example 4

Calculate the equivalent weights for the following salts: Na2O, Na2CO3, Al2(SO4)3.

Atomic weights for O = 16, C = 12, Na = 23, S = 32, Al = 26.98.

molar mass

equivalent weight for Na2O = ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ

no. of metal ions × oxidation no.

(23 × 2) + 16

gram/equivalent 31 = ـــــــــــــــــــــــــــــــــــــــ =

2 × 1

(23 × 2) + 12 + (16 × 3)

for Na2CO3 = 53 = ـــــــــــــــــــــــــــــــــــــــــــــــــ

2 × 1

(26.98 × 2) + [32 + (16 × 4)] 3

for Al2(SO4)3 = 56.99 = ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ

2 × 3

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For salts in precipitation reactions:

The equivalent weight of salts in precipitation reactions is the weight of substance

in gram that precipitates quantity equivalent to quantity of 1 gram of hydrogen, or the

equivalent weight of another substance in the same reaction.

molar mass

equivalent weight for salt = ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ

sum of oxidation numbers of the part

participates in precipitate formation

Example 5

Calculate the equivalent weight for the substances participate in the reaction of AgCl

precipitation. Atomic weights for Ag = 108, N = 14, O = 16, Na = 23, Cl = 35.5.

3AgCl + NaNO + NaCl 3AgNO

molar mass for AgNO3

equivalent weight for AgNO3 = ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ

oxidation number of Ag

108 + 14 + (16 × 3)

ـــــــــــــــــــــــــــــــــــــــــــــــ = = 170 gram/equivalent

1

molar mass for NaCl

equivalent weight for NaCl = ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ

oxidation number of Cl-

35.5 + 23

58.5 = ــــــــــــــــــــــــــــــ =

1

For oxidation-reduction agents:

The oxidation-reduction reactions involve transfer of electrons from one substance

to another.

molar mass equivalent weight for oxidation = ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ

or reduction agent the difference between oxidation numbers

13

Example 6

Calculate the equivalent weight for FeSO4 and KMnO4 in the following reaction,

which permanganate solution oxidizes iron sulphate. Atomic weights for Fe = 56, S =

32, O = 16, K = 39, Mn = 55.

MnO4- + 5Fe2+ + 8H+ Mn2+ + 5Fe3+ + 4H2O

The ferrous oxidized to ferric:

Fe2+ Fe3+ + e- (oxidation)

the difference between oxidation numbers = +3 – (+2) = +1

molar mass for FeSO4

equivalent weight for FeSO4 = ــــــــــــــــــــــــــــــــــــــــــــــــــــــــ

1

152

gram/equivalent 152 = ـــــــــــــــــــ =

1

the manganese gain 5 electrons:

Mn7+ + 5e- Mn2+ (reduction)

the difference between oxidation numbers = +7 – (+2) = 5

molar mass for KMnO4

equivalent weight for KMnO4 = ــــــــــــــــــــــــــــــــــــــــــــــــــــــــ

5

158

31.6 = ـــــــــــــــــــ =

5

14

1.8 Methods Expressing Solutions concentration

g solute

1- Mass percentage w/w% = 100 × ــــــــــــــــــــــــــــ

g solution

5% solution of NaCl means that 5 g of NaCl dissolved in 95 g water.

Example 7

Calculate the mass percentage of 200 g solution contains 25 g sodium sulphate.

g solute

mass percentage w/w% = 100 × ــــــــــــــــــــــــــــ

g solution

25

%12.5 = 100 × ــــــــــــــــــــــــــــ =

200

Example 8

Calculate the mass percentage for a solution prepared by dissolving 15 g of AgNO3 in

100 cm3 water. The density of water is 1 g/cm3.

weight of solvent = volume × density

= 100 cm3 × 1 g/cm3

= 100 g

weight of solution = 100 g solvent + 15 g solute

= 115 g

g solute


g solution

15

15

100 × ــــــــــــــــــــــــــــ =

115

= 13.04%

ml solute

2- Volume percentage v/v% = 100 × ــــــــــــــــــــــــــــ

ml solution

10% solution of alcohol means that 10 ml alcohol is added to enough solvent in order

to reach 100 ml volume (addition of 90 ml solvent).

Example 9

10 g of organic solvent (density 1.5 g/cm3) was added to 90 g water, the density of the

solution become 1.1 g/cm3. Calculate the v/v% and the w/w% concentrations of the

organic substance in the solution.

weight of solution = 90 g + 10 g = 100 g

10 g


100 g

= 10%

weight 100

volume of solution = 90.90 = ـــــــــــــــ = ـــــــــــــــــــــ ml

density 1.1

weight 10

volume of solute = 6.67 = ـــــــــــــــ = ـــــــــــــــــــــ ml

density 1.5

6.67 ml

volume percentage v/v% = ــــــــــــــــــــــــــــ × 100 = 7.3%

90.90 ml

16

g solute

3- Mass/volume percentage w/v% = 100 × ــــــــــــــــــــــــــــ

ml solution

Example 10

Calculate the weight of sodium chloride salt in 500 ml solution of a 0.85 %w/v

concentration.

g solute

w/v% = 100 × ــــــــــــــــــــــــــــ

ml solution

g

g 4.25 = 100 × ــــــــــــــــ = 0.85

500

mg solute

4- Parts per million (ppm) = ـــــــــــــــــــــــ

kg solvent

and there are:

g solute

Parts per thousand (ppt) = ـــــــــــــــــــــــ

kg solvent

µg solute

Parts per billion (ppb) = ـــــــــــــــــــــــ

kg solvent

Example 11

A weight of a sample 345 g contains 3 mg Hg, what is the concentration of Hg in the

sample in ppm?

mg solute

parts per million (ppm) = ـــــــــــــــــــــــ

kg solvent

3

(ppm) = 7.35 = ـــــــــــــــــــــــ ppm

0.345

17

Example 12

A sample contains 4.8 parts per billion arsine, if the weight of the sample is 525 g,

how many µg arsine present in the sample?

µg solute

parts per billion (ppb) = ـــــــــــــــــــــــ

kg solvent

µg arsine

ـــــــــــــــــــــــ = 4.8

0.525

µg arsine = 2.52

5- Molar concentration (M)

It is the number of molar weights of the solute in 1 liter of solvent.

no. of molar weights of solute

Molarity (M) = ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ

volume of solution in liter

weight of substance in g

no. of molar weights = ــــــــــــــــــــــــــــــــــــــــــــــــــــــــ

molar mass


ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ

molar mass

= Molar concentration (M)ـــــــــــــ ــــــــــــــــــــــــــــــــــــــــــــــــــــــــ

volume of solution in ml

ـــــــــــــــــــــــــــــــــــــــــــــــــــــــ

1000

weight of substance in g × 1000

ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ =

molecular weight × volume of solution in ml

volume ml

weight of substance = M × molecular weight × ــــــــــــــــــــــــــ

1000

18

Example 13

Calculate the molar concentration (M) of a solution prepared by dissolving 29.35 g

of NaCl in 200 ml water. Atomic weights for Na = 22.99, Cl = 35.45.


M = ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ


29.35 × 1000

molar 2.5 = ـــــــــــــــــــــــــــــــــــــــ =

200 × 58.44

Example 14

Calculate the weight of AgNO3 needed to prepare 500 ml solution of a concentration

0.1250 M. Molecular weight of AgNO3 is 169.9.


M = ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ


volume ml

weight of AgNO3 = M × molecular weight × ــــــــــــــــــــــــــ

1000

500

weight of AgNO3 = 0.125× 169.9× ــــــــــــــــــــــــــ

1000

= 10.62 g

19

6- Normal concentration (N)

It is the number of equivalent weights of the solute dissolved in liter of the solvent.

no. of equivalent weights of solute

Normality (N) = ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ

volume of solution in liter


no. of equivalent weights = ــــــــــــــــــــــــــــــــــــــــــــــــــــــــ

equivalent weight


ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ

equivalent weight

= Normal concentration (N)ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ

volume of solution in ml

ـــــــــــــــــــــــــــــــــــــــــــــــــــــــ

1000


ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ =

equivalent weight × volume of solution in ml

volume ml

weight of substance = N × equivalent weight × ــــــــــــــــــــــــــ

1000

Example 15

Calculate the number of grams of Na2SO4 needed to prepare 200 ml solution of 0.5 N

concentration. Equivalent weight of Na2SO4 = 71.

volume ml

weight of substance = N × equivalent weight × ــــــــــــــــــــــــــ

1000

200

weight of Na2SO4 = 0.5 × 71 × 7.1 = ــــــــــــــــــــ g

1000

20

Example 16

Calculate the molar (M) and normal (N) concentrations for a solution prepared by

dissolving 10.6 g from sodium carbonate Na2CO3 in a liter of the solution. Molecular

weight of Na2CO3 = 106.

weight

no. of moles of Na2SO4 = ــــــــــــــــــــــــــــــــــــــ

molecular weight

10.6

0.1 = ــــــــــــــــــــــ =

106

no. of moles of solute 0.1

Molarity (M) = ــــــــــــــ = ــــــــــــــــــــــــــــــــــــــــــــــــــــــــ = 0.1

volume of solution in liter 1

weight 10.6

no. of equivalent weights of Na2SO4 = ـــــــــ = ــــــــــــــــــــــــــــــــ

equivalent weight 106/2

= 0.2

no. of equivalent weights 0.2

Normality (N) = ـــــــــــــــــــــــــــــــــــــــــــــــــــــــ 0.2 = ـــــــــــ =

volume of solution in liter 1

* Relationship between Molarity (M) and Normality (N)

N = M × no. of equivalents

Example 17

Calculate the molar (M) concentration of H3PO4 solution of 0.250 N, to produce

phosphate ion PO43-.

N = M × no. of equivalents

0.25 = M × 3

M = 0.0833 Molar

21

7- Molal concentration (m)

It is the number of molar masses (moles) of the dissolved substance in 1000 g of

the solvent, whatever is the total volume of the solution.


Molal concentration (m) = ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ

molecular weight × weight of solvent in g

Example 18

Calculate the molal (m) concentration m for ethanol in a solution prepared by

dissolving 92.2 g ethanol in 500 g water. Molecular weight for ethanol = 46.1.


Molal concentration (m) = ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ

molecular weight × weight of solvent in g

92.9 × 1000

ـــــــــــــــــــــــــــــــــ = = 4 m

46.1 × 500

* The mole fraction

The mole fraction for solvent is the number of moles of solvent relative to the total

number of moles, and the mole fraction for solute is the number of moles of solute

relative to the total number of moles. If we multiply the mole fraction by 100 the

product is mole percent.

Example 19

Calculate the mole fraction for ethanol C2H5OH and water in a solution prepared by

dissolving 13.80 g of ethanol in 27 g water. Atomic weight for O=16, C=12, H=1.

weight 13.80

no. of moles of ethanol = 0.30 = ـــــــــــــــــ = ـــــــــــــــــــــــــــــــ mole

molecular weight 46

22

weight 27

no. of moles of water = 1.50 = ـــــــــــــــــ = ـــــــــــــــــــــــــــــــ mole

molecular weight 18

the total number of moles = 0.30 + 1.50 = 1.80 mole

moles of ethanol 0.30

mole fraction for ethanol = 0.167 = ـــــــــ = ـــــــــــــــــــــــــــــــــــــ

total no. of moles 1.80

moles of water 1.50

mole fraction for water = 0.833 = ـــــــــ = ـــــــــــــــــــــــــــــــــــــ

total no. of moles 1.80

The highest value for mole fraction is 1, so it is possible to calculate mole fraction for

water by determining mole fraction for ethanol:

mole fraction for water = 1 – mole fraction for ethanol

= 1- 0.167 = 0.833

* Solutions Normality

D × % × 1000

The concentration of an acid can be calculated as follow: N = ــــــــــــــــــــــــــــــــــــــــــــــ

equivalent weight × 100

D = density.

To dilute a solution:

N1 × V1 = N2 × V2

before dilution after dilution

24

2.1 Introduction

Every physical measurement is subject to a degree of uncertainty that, at best, can

be decreased only to an acceptable level. The determination of the magnitude of this

uncertainty is often a difficult task that requires additional effort, ingenuity, and good

judgment on the part of the scientist.

A direct relationship exists between the accuracy of an analytical measurement and

the time and effort spend.

2.2 The Mean and Median

The mean, arithmetic mean, and average ( ) are synonymous terms for the

numerical value obtained by dividing the sum of a set of replicate measurements by

the number of individual results in the set.

The median of a set is that result about which all others are equally distributed, half

being numerically greater and half being numerically smaller. If the set consists of an

odd number of measurements, selection of the median may be made directly; for a set

containing an even number of measurements, the average of the central pair is taken.

Example 1

Calculate the mean and the median for 10.06, 10.20, 10.08, 10.10.

10.06 + 10.20 + 10.08 + 10.10

mean = ( ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ = ( = 10.11

4

Since the set contain an even number of measurements, the median is the average

of the middle pair:

10.08 + 10.10

median = 10.09 = ــــــــــــــــــــــــــــــــــ

2

Ideally, the mean and the median should be numerically identical; more often than not,

however, this condition is not realized, particularly when the number of measurements

in the set is small.

25

2.3 Precision

The term precision is used to describe the reproducibility of results. It can be

defined as the agreement between the numerical values of two or more measurements

that have been made in an identical fashion. Several methods exist for expressing the

precision of data.

* Absolute methods for expressing precision

The deviation from the mean ( ) is a common method for describing

precision and is simply the numerical difference, without regard to sign, between an

experimental value and the mean for the set of data that include the value. To

illustrate, suppose that a chloride analysis yielded the following results:

Sample Percent

Chloride

Deviation from Mean

( )

Deviation from

Median

x1 24.39 0.077 0.03

x2 24.19 0.123 0.17

x3 24.36 0.047 0.00

3[72.94 3[0.247 3[0.20

( ) = 24.313 = 24.31 Avg = 0.082 = 0.08 Avg = 0.067 = 0.07

w = xmax – xmin = 24.39 – 24.19 = 0.20

ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ

Precision can also be reported in terms of deviation from the median. In the

preceding example, deviations from 24.36 would be recorded, as shown in the last

column of the table. The spread or range (w) in a set of data is the numerical

difference between the highest and lowest result and also a measure of precision. In

the previous example, the spread would be 0.20% chloride.

Two other measures of precision are the standard deviation and the variance.

26

* Relative precision

We have thus far considered the expression of precision in absolute terms. It is

often more convenient, however, to indicate the precision relative to the mean (or the

median) in terms of percentage or as parts per thousand. For example, for sample x1,

0.077 × 100

relative deviation from mean = ــــــــــــــــــــــــــــــــــ = 0.32 0.3%

24.31

Similarly, the average deviation of the set from the median can be expressed as:

0.067 × 1000

relative average deviation from median = ــــــــــــــــــــــــــــــ = 3 ppt

24.36

2.4 Accuracy

The term accuracy denotes the nearness of a measurement to its accepted value

and is expressed in terms of error. Accuracy involves a comparison with respect to a

true or accepted value; in contrast, precision compares a result with other

measurements in the same way.

The accuracy of a measurement is often described in terms of absolute error,

which can be defined as:

E = xi – xt

The absolute error E is the difference between the observed value xi and the accepted

value xt. The value may itself be subjected to considerable uncertainty.

Returning to the previous example, suppose that the accepted value for the

percentage of chloride in the sample is 24.36%. The absolute error of the mean is thus

24.31 – 24.36 = - 0.05% chloride; here we ordinarily retain the sign of the error to

indicate whether the result is high or low.

Often a more useful quantity than the absolute error is the relative error expressed

as a percentage or in part per thousand of the accepted value. Thus, for the chloride

analysis we have been considering,

27

0.05 × 100

relative error = 0.2 = 0.21 - = ــــــــــــــــــــــــــــــــــ%

24.36

0.05 × 1000

relative error = 2 - = 2.1 - = ــــــــــــــــــــــــــــــــــ ppt

24.36

2.5 Precision and Accuracy of Experimental Data

The precision of a measurement is readily determined by replicate experiments

performed under identical conditions. Unfortunately, an estimate of the accuracy is not

equally available because this quantity requires sure knowledge of the vary

information that is being sought, namely, the true value. It is tempting to ascribe a

direct relationship between precision and accuracy. The danger of this approach is

illustrated in the following figure:

2.6 Classification of Errors

The uncertainties that arise in a chemical analysis, and that are responsible for the

behavior illustrated in the last figure, may be classified into two broad categories,

depending upon their origin.

28

* Determinate errors: those that have a definite value which can (in principle, if not

in practice) be measured and accounted for.

*Indeterminate errors: result from extending a system of measurement to its

maximum. These errors cannot be positively identified and do not have a definite

measurable value; instead, they fluctuate in a random manner.

2.7 Types of Determinate Errors

1- Personal errors

These errors are the result of the ignorance, carelessness, prejudices, or physical

limitations of the experimenter. For example, they may arise from the use of an

improper technique in transferring a sample, from disregard of temperature corrections

for a measuring device, from over- or underwashing a precipitate, or from transposing

numbers when recording an experimental measurement.

2- Instrumental errors

Instrumental errors are attributable to imperfections in the tools with which the

analyst works or to the effects of environmental factors upon these tools. For example,

volumetric equipment such as burets, pipets, and volumetric flasks frequently deliver

or contain volumes slightly different from those indicated by their graduations,

particularly when they are employed at temperatures which differ significantly from

the temperature at which they were calibrated. Calibration at the proper temperature

will obviously eliminate this type of determinate error.

3- Method errors

Determinate errors are often introduced from nonideal chemical behavior of the

reagents and reactions upon which the analysis is based. Such sources of nonideality

include the slowness of some reactions, the incompleteness of others, the lack of

stability of some species, the nonspecificity of most reagents, and the possible

occurrence of side reactions which interfere with the measurement process. For

example, in a gravimetric analysis the chemist is confronted with the problem of

29

isolating the element to be determined as a solid of the greatest possible purity. If he

fails to wash it sufficiently, the precipitate will be contaminated with foreign

substances and have a spuriously high weight. On the other hand, sufficient washing to

remove these contaminants may cause weighable quantities to be lost owing to the

solubility of the precipitate; here, a negative determinate error will result.

2.8 Effect of Determinate Error upon the Results of an Analysis

Determinate errors generally fall into either of two categories, constant or

proportional. The magnitude of a constant error stays essentially the same as the size

of the quantity measured is varied. Proportional errors increase or decrease according

to the size of the sample taken for analysis.

* Constant errors. For any given analysis, a constant error will become more serious

as the size of the quantity measured decreases. This problem is illustrated by the

solubility losses that attend the washing of a precipitate.

Example 2

Suppose that directions call for washing of the precipitate with 200 ml of water, and

that 0.50 mg is lost in this volume of wash liquid. If 500 mg of precipitate are

involved, the relative error due to solubility loss will be:

- (0.50 × 100)

%0.1 - = ــــــــــــــــــــــــــــــــــــ

500

Loss of the same quantity from 50 mg of precipitate will result in a relative error of –

1.0%.

The amount of reagent required to bring about the color change in a volumetric

analysis is another example of constant error. This volume, usually small, remains the

same regardless of the total volume of reagent required. Again, the relative error will

be more serious as the total volume decreases. Clearly, one way of minimizing the

effect of constant error is to use as large a sample as is consistent with the method at

hand.

30

* Proportional errors. Interfering contaminants in the sample, if not eliminated in

some manner, will cause an error of the proportional variety. For example, a method

widely employed for the analysis of copper involves reaction of the copper(II) ion

with potassium iodide; the quantity of iodine produced in the reaction is then

measured. Iron(III), if present, will also liberate iodine from potassium iodide. Unless

steps are taken to prevent this interference, the analysis will yield erroneously high

results for the percentage of copper since the iodine produced will be a measure of the

sum of the copper and iron in the sample. The magnitude of this error is fixed by the

fraction of iron contamination and will produce the same relative effect regardless of

the size of sample taken for analysis. If the sample size is doubled, for example, the

amount of iodine librated by both the copper and the iron contaminant will also be

doubled.

2.9 Effects of Indeterminate Error

As suggested by its name, indeterminate error arises from uncertainties in a

measurement that are unknown and cannot be controlled by the scientist. The effect of

such uncertainties is to produce a scatter of results for replicate measurements.

2.10 The standard deviation

The absolute standard deviation, s, describes the spread of individual measurements

about the mean and is given as:

where xi is one of n individual measurements, and is the mean. Frequently, the

relative standard deviation, sr, is reported. The percent relative standard deviation is

obtained by multiplying sr by 100%.

31

Example 3

Calculate the standard deviation s for a subset consisting of the following values of the

replicate measurements from the calibration of a 10 ml pipette.

Trial Vol. water delivered, ml

1 9.990

2 9.986

3 9.973

4 9.983

5 9.980

xi

9.990 7.6 × 10-3 57.8 × 10-6

9.986 3.6 × 10-3 13.0 × 10-6

9.973 9.4 × 10-3 88.4 × 10-6

9.983 0.6 × 10-3 0.4 × 10-6

9.980 2.4 × 10-3 5.8 × 10-6

5 49.912

9.9824 =

We can also calculate sr:

32

2.11 The Variance

The variance is another common measure of spread; it is the square of the standard

deviation. The standard deviation, rather than the variance, is usually reported because

the units for standard deviation are the same as that for the mean value.

To calculate the variance of the last example:

Variance = s2 = (0.006)2 = 0.000036

2.12 Rejection of Data

When a set of data contains an outlying result that appears to differ excessively

from the average (or the median), the decision must be made to retain or to reject it.

Of the numerous statistical criteria suggested to aid in deciding whether to retain

or reject a measurement, the Q test is to be preferred. Here the difference between the

questionable result and its nearest neighbor is divided by the spread of the entire set.

The resulting ratio, Q, is then compared with rejection values that are critical for a

particular degree of confidence. Table 2-1 provides critical values of Q at the 90%

confidence level. Q = rejection quotient, Qcrit = critical Q-value, Qexp = experimental Q-

value.

Table 2-1 Critical Values for Rejection Quotient Q

Number of

Observations

Qcrit (90% confidence)

Reject if Qexp ˃

2 -

3 0.94

4 0.76

5 0.64

6 0.56

7 0.51

8 0.47

9 0.44

10 0.41

33

Example 4

The analysis of a calcite sample yielded CaO percentage of 55.95, 56.00, 56.04, 56.08,

and 56.23, respectively. The last value appears anomalous; should it be retained or

rejected?

The difference between 56.23 and 56.08 is 0.15. The spread (56.23 – 55.95) is 0.28.

Thus,

0.15

Qexp = 0.54 = ــــــــــــــــــــــ

0.28

For five measurements, Qcrit is 0.64. Since 0.54 ˂ 0.64, retention is indicated.

35

3.1 Introduction

A gravimetric analysis is based upon the measurement of the weight of a substance

of known composition that is chemically related to the analyte. Two types of

gravimetric methods exist.

1- Precipitation methods: The species to be determined is caused to react chemically

with a reagent to yield a product of limited solubility; after filtration and other suitable

treatment, the solid residue of known chemical composition is weighed.

2- Volatilization methods: The substance to be determined is separated as a gas from

the remainder of the sample; here the analysis is based upon the weight of the

volatilized substance or upon the weight of the nonvolatile residue. Precipitation

methods are more frequently encountered than methods involving volatilization, so we

concentrate on the precipitation method.

3.2 Calculation of Results from Gravimetric Data

A gravimetric analysis requires two experimental measurements: specifically, the

weight of sample taken and the weight of a product of known composition derived

from the sample. Ordinarily these data are converted to a percentage of analyte by a

simple mathematical manipulation.

If A is the analyte, we may write:

weight of A

% A = ــــــــــــــــــــــــــــــــــ × 100 (3-1)

weight of sample

Usually the weight of A is not measured directly. Instead, the species that is actually

isolated and weight either contains A or can be chemically related to A. In either case,

a gravimetric factor is needed to convert the weight of the precipitate to the

corresponding weight of A. The properties of this factor are conveniently

demonstrated with examples.

36

Example 1

How many grams of Cl are contained in a precipitate of AgCl that weights 0.204 g?

Atomic weights for Ag = 107.86, Cl = 35.45.

We write the equation:

AgNO3 + NaCl AgCl↓ + NaNO3

Cl- AgCl

35.45 ≡ 143.3

x 0.204

35.45

x = 0.204 × 0.0505 = ـــــــــــــــــــــــ g

143.3

35.45

The value ـــــــــــــــ is called the gravimetric factor (GF).

143.3

Example 2

To what weight of AlCl3 would 0.204 g of AgCl correspond? Atomic weights for Ag

= 107.86, Cl = 35.45, Al = 26.98.

We know that each AlCl3 yield three AgCl. Therefore,

AlCl3 ≡ 3AgCl

133.3 ≡ 3 × 143.3

x 0.204

133.3


3 × 143.3

37

Example 3

What weight of Fe2O3 can be obtained from 1.63 g of Fe3O4? What is the gravimetric

factor for this conversation? Atomic weights for Fe = 55.84, O = 16.

It is necessary to assume that all Fe in the Fe3O4 is transformed into Fe2O3 and ample

oxygen is available to accomplish this change. That is:

2Fe3O4 + [O] = 3Fe2O3

2Fe3O4 ≡ 3Fe2O3

2 × 231.5 ≡ 3 × 159.7

1.63 x

3 × 159.7

x = 1.63 × 1.69 = 1.687 = ـــــــــــــــــــــــ g

2 × 231.5

3 × 159.7

gravimetric factor 1.035 = ـــــــــــــــــــــــــــ

2 × 231.5

Example 4

A 0.703 g sample of a commercial detergent was ignited at a red heat to destroy the

organic matter. The residue was then taken up in hot HCl which converted the P to

H3PO4. The phosphate was precipitated as MgNH4PO4.6H2O by addition of Mg2+

followed by aqueous NH3. After being filtered and washed, the precipitate was

converted to Mg2P2O7 by ignition at 1000oC. This residue weighed 0.432 g. Calculate

the percent P in the sample. Atomic weights for P = 30.97, Mg = 24.3, O = 16.

P ≡ Mg2P2O7

30.97 ≡ 222.54

x 0.432

38

2 × 30.97


222.54

0.120

% P = 17.1 = 100 × ـــــــــــــــــــ%

0.703

Example 5

At elevated temperatures sodium oxalate is converted to sodium carbonate:

Na2C2O4 Na2CO3 + CO

Ignition of a 1.3906 g sample of impure sodium oxalate yielded a residue weighing

1.1436 g. Calculate the percentage purity of the sample. Atomic weights for Na =

22.98, C = 12, O = 16.

Here it must be assumed that the difference between the initial and final weights

represents the carbon monoxide evolved during the ignition; it is this weight loss that

forms the basis for the analysis. From the equation for the process, we see that:

1.3906 – 1.1436 = 0.247 weight of CO

Na2C2O4 ≡ CO

133.96 ≡ 28

x 0.247

133.96

x = 0.247 × 1.181 = ــــــــــــــــــــ g

28

1.181

% Na2C2O4 = 84.97 = 100 × ـــــــــــــــــــ%

1.3906

39

Example 6

Calculate the percentage of SO3 in a sample of gypsum of a 0.7560 g weight, produced

0.9875 g of BaSO4 precipitate. Atomic weights for Ba = 137.34, O = 16, S = 32.

BaSO4 ≡ SO3

233.34 ≡ 80

0.9875 x

80

x = 0.9875 × 0.3385 = ــــــــــــــــــــg

233.34

0.3385

% SO3 = 44.77 = 100 × ـــــــــــــــــــ%

0.7560

3.3 Properties of Precipitates and Precipitating Reagents

The ideal precipitating reagent for a gravimetric analysis would react specifically

with the analyte to produce a solid that would:

(1) have a sufficiently low solubility so that losses from that source would be

negligible.

(2) be readily filtered and washed free of contaminants.

(3) be unreactive and of known composition after drying or, if necessary, ignition.

Few precipitates or reagents possess all these desirable properties; thus, the

chemist frequently finds it necessary to perform analyses using a product or a reaction

that is far from ideal.

3.4 Filterability and Purity of Precipitates

Both the ease of filtration and the ease of purification are influenced by the particle

of the solid phase. The relationship between particle size and ease of filtration is

straightforward, coarse precipitates being readily retained by porous media which

permit rapid filtration. Finely divided precipitates require dense filters, low filtration

40

rates result. The effect of particle size upon purity of a precipitate is more complex. A

decrease in soluble contaminants is found to accompany an increase in particle size.

In considering the purity of precipitates we shall use the term coprecipitation,

which describes those processes by which normally soluble components of a solution

are carried down during the formation of a precipitate.

3.5 Factors That Determine the Particle Size of Precipitates

Enormous variation is observed in the particle size of precipitates, depending upon

their chemical composition and the conditions leading to their formation.

* Colloidal suspensions, the individual particles of which are so small as to be

invisible to the naked eye (10-6 to 10-4 mm in diameter). These particles show no

tendency to settle out from solution, nor are they retained upon common filtering

media.

* Particles with dimensions on the order of several tenths of a millimeter. The

dispersion of such particles in the liquid phase is called a crystalline suspension. The

particles of a crystalline suspension tend to settle out rapidly and are readily filtered.

No sharp discontinuities in physical properties occur as the dimensions of the

particles containing the solid phase increase from colloidal to those typical of crystals.

Indeed, some precipitates possess characteristics intermediate between these types.

The particle size of the solid that forms is influenced in part by experimental

variables as the temperature, the solubility of the precipitate in the medium in which it

is being formed, reactant concentrations, and the rate at which reagents are mixed. The

particle size is related to a single property called relative supersaturation, where:

Q - S

relative supersaturation = (2-3) ـــــــــــــــــــــــــــــ

S

Q is the concentration of the solute at any instant, and S is its equilibrium solubility.

During formation of sparingly soluble precipitate, each addition of precipitating

reagent causes the solution to be momentarily supersaturated (that is Q ˃ S). Under

41

most circumstances, this unstable condition is relived, usually after a brief period, by

precipitate formation. Experimental evidence suggests, however, that the particle size

of the resulting precipitate varies inversely with the average of relative supersaturation

that exists after each addition of reagent. Thus, when (Q – S) / S is large, the

precipitate tends to be colloidal; when this parameter is low on the average, a

crystalline solid results.

3.6 Mechanics of Precipitate Formation

The effect of relative supersaturation on particle size can be rationalized by

postulating two precipitation processes, nucleation and particle growth. The particle

size of a freshly formed precipitate is governed by the extent to which one of these

steps predominates over the other.

Nucleation is a process whereby some minimum number of ions or molecules

(perhaps as few as four or five) units to form a stable second phase. Further

precipitation can occur either by formation of additional nuclei or by deposition of

solid on the nuclei that are already present (particle growth). If the former

predominates, a precipitate containing a large number of small particles results; if

growth predominates, a smaller number of large particles will be produced.

The rate of nucleation is believed to increase enormously with increasing

relativesupersaturation. In contrast, the rate of particle growth is only moderately

enhanced by high relative supersaturations.

At low relative supersaturations, growth predominates. When the supersaturation is

great, the exponential nature of nucleation may cause this process to occur. These

effects are illustrated in Figure 3-1.

42

Figure 3-1 Effect of Relative Supersaturation on Precipitation Processes

3.7 Experimental Control of Particle Size

Experimental variables that minimize supersaturation and thus lead to crystalline

precipitate include:

Elevated temperature (to increase S).

Dilute solutions (to minimize Q).

Low addition of the precipitating agent and

Good stirring (also to lower the average value of Q).

The particle size of precipitates with solubilities that are pH-dependent can often

be enhanced by increasing S during precipitation. For example, large, easily filtered

crystals of calcium oxalate can be obtained by forming the bulk of the precipitate in a

somewhat acidic environment in which the salt is moderately soluble. The

precipitation is then completed by slowly adding aqueous ammonia until the pH is

sufficiently high for quantitative removal of the calcium oxalate; the additional

precipitate produced during this step forms on the solid.

43

3.8 Colloidal Precipitates

Individual colloidal particles are so small that they are not retained on ordinary

filtering media; furthermore, Brownian motion prevents their settling out of solution

under the influence of gravity. Fortunately, however, the individual particles of most

colloids can be coagulate or agglomerate to give a filterable, noncrystalline mass that

rapidly settles out from a solution.

* Coagulation of colloids

Three experimental measures induce the coagulation process are, heating,

stirring, and adding an electrolyte to the medium. To understand the effectiveness

of these measures, we need to account for the stability of a colloidal suspension.

The individual particles in a typical colloid bear either a positive or a negative

charge as a consequence of adsorption of cations or anions on their surfaces.

Adsorption of ions upon an ionic solid has the normal bonding forces that are

responsible for crystal growth. When a silver ion adsorbed on the surface of a silver

chloride particle, negative ions are attracted to this site by the same forces that hold

chloride ions in the silver chloride lattice. Chloride ions on the surface exert an

analogous attraction for cations in the solvent.

Thus, a silver chloride particle will be positively charged in a solution containing

an excess of silver ions, and it will have a negative charge in the presence of excess

chloride ion for the same reason.

Figure 3-2 illustrates schematically a colloidal silver chloride particle in a solution

containing an excess of silver ions. Attached directly to the solid surface are silver

ions in the primary adsorption layer. Surrounding the charge particle is a region of

solution called the counter-ion layer, within which there exists an excess of negative

ions sufficient to balance the charge of the adsorbed positive ions on the particle

surface. The counter-ion layer forms as the result of electrostatic forces.

44

Figure 3-2 Colloidal AgCl Particle Suspended in a Solution of AgNO3

The solution outside the secondary adsorption layer remains electrically neutral.

Coagulation cannot occur if the secondary adsorption layer is too thick because the

individual particles of AgCl are unable to approach one another closely enough. The

primarily adsorbed silver ions and the negative counter-ion layer constitute an electric

double layer that imparts stability to the colloidal suspension. As colloidal particles

approach one another, this double layer exerts an electrostatic repulsive force that

prevents particles from colliding and adhering.

Coagulation of a colloidal suspension can often be brought about by a short period

of healing, particularly if accompanied by stirring. Heating deceases the number of

adhered ions and thus the thickness of the double layer.

45

An even more effective way to coagulate a colloid is to increase the electrolyte

concentration of the solution. If we add a suitable ionic compound to a colloidal

suspension, the concentration of counter-ions increases in the vicinity of each particle.

As a result, the volume of solution that contains sufficient counter-ions to balance the

charge of the primary adsorption layer decreases.

* Coprecipitation in coagulated colloids

Adsorption is the principal type of coprecipitation that affects coagulated colloids;

other types are encountered with crystalline solids.

A coagulated colloid consists of irregularly arranged particles which form a

loosely packed, porous mass. Within this mass, large internal surface areas remain in

contact with the solvent phase. Adhering to these surfaces will be most of the

primarily adsorbed ions which were on the uncoagulated particles. Even though the

counter-ion layer surrounding the original colloidal particle is properly considered to

be part of the solution, it must be recognized that sufficient counter ions to impart

electrical neutrality must accompany the particle through the processes of coagulation

and filtration.

* Peptization of colloids

Peptization is the process whereby a coagulated colloid reverts to its original

dispersed state. Peptization occurs when pure water is used to wash such a precipitate.

Washing is not particularly effective in dislodging adsorbed contaminants; it does tend

to remove the electrolyte responsible for coagulation from the internal liquid in

contact with the solid. As the electrolyte is removed, the counter-ion layers increase

again in volume. The repulsive forces responsible for the colloidal state are thus

reestablished, and particles detach themselves from the coagulated mass. The washing

becomes cloudy as the freshly dispersed particles pass through the filter.

This problem is commonly resolved by washing the agglomerated colloid with a

solution containing a volatile electrolyte which can subsequently be removed from the

46

solid by heating. For example, silver chloride precipitates are ordinarily washed with

dilute nitric acid. The washed precipitate is heavily contaminated by the acid, but no

harm results since the nitric acid is removed when the precipitate is dried at 110oC.

3.9 Crystalline Precipitates

In general, crystalline precipitates are more easily handled than coagulated

colloids. The particle size of crystalline solids can be improved by keeping the relative

supersaturation low during the period in which the precipitate is formed. From

Equation 3-2, it is apparent that minimizing Q or maximizing S, or both, will

accomplish this purpose.

The use of dilute solutions and the slow addition of precipitating agent with good

mixing tend to reduce the supersaturation in the solution. Ordinarily, S can be

increased by precipitating from hot solution.

* Types of impurities in crystalline precipitate

The specific area of crystalline precipitates is relatively small; consequently,

coprecipitation by direct adsorption is negligible. However, other forms of

coprecipitation, which involve incorporation of contamination within the interior of

crystals, may cause serious errors. There are two types of coprecipitation:

1- Inclusion: Interfering ions whose size and charge are similar to a lattice ion may

substitute into the lattice structure by chemical adsorption. The impurities are

randomly distributed, in the form of individual ions or molecules, throughout the

crystal. Inclusions are difficult to remove since the included material is chemically

part of the crystal lattice. The only way to remove included material is through

reprecipitation.

2- Occlusion: It involves a nonhomogeneous distribution of impurities, consisting of

numerous ions or molecules of the contaminant, within imperfections in the crystal

lattice.

Occlusion occurs when whole droplets of solution containing impurities are

trapped and surrounded by a rapidly growing crystal. Because the contaminants are

47

located within the crystal, washing does little to decrease their amount. A lower

precipitation rate may significantly lessen the extent of occlusion by providing time

for the impurities to escape before they become entrapped. Digestion of the precipitate

for as long as several hours is even more effective in eliminating contamination by

occlusion.

* Digestion of crystalline precipitates

The heating of crystalline precipitates (without stirring) for some time after

formation frequently yields a product with improved purity and filterability. The

improvement in purity results from the solution and recrystallization that occur

continuously and at an enhanced rate at elevated temperatures. During these processes,

many pockets of imperfection become exposed to the solution; the contaminant is thus

able to escape from the solid and more perfect crystals result.

Solution and recrystallization during digestion are probably responsible for the

improvement in filterability as well. Bridging between adjacent particles occur to yield

larger crystalline aggregates which are more easily filtered.

3.10 Direction of Coprecipitation Errors

Coprecipitated impurities may cause the results of an analysis to be either too high

or too low. If the contaminant is not a compound of the ion being determined, positive

errors will always result. Thus, a positive error will be observed when colloidal silver

chloride adsorbs silver nitrate during a chloride analysis. On the other hand, where the

contaminant contains the ion being determined, either positive or negative errors may

be observed. In the determination of barium ions by precipitation as barium sulfate, for

example, occlusion of barium salts occurs. If the occluded contaminant is barium

nitrate, a positive error will be observed, since this compound has a greater formula

weigh than the barium sulfate. If barium chloride were the contaminant, however, a

negative error would arise because its formula weight is less than that of the sulfate

salt.

48

3.11 Drying and Ignition of Precipitates

After separating the precipitate from its supernatant solution the precipitate is dried

to remove any residual traces of rinse solution and any volatile impurities. The

temperature and method of drying depend on the method of filtration, and the

precipitate’s desired chemical form. A temperature of 110 °C is usually sufficient

when removing water and other easily volatilized impurities. A conventional

laboratory oven is sufficient for this purpose. The temperature required to produce a

suitable product varies from precipitate to precipitate.

3.12 Applications of Gravimetric Method

* Inorganic precipitating agents

Table 3-1 lists some of the common inorganic precipitating agents. These reagents

typically cause formation of a slightly soluble salt or a hydrous oxide.

The weighing form is either the salts itself or else an oxide. The lack of specificity

of most inorganic reagent is clear from the many entries in the table.

Table 3-1 Some Inorganic Precipitating Agents

49

* Reducing reagents

Table 3-2 lists several reagents that convert the analyte to its elemental form for

weighing.

Table 3-2 Some Reducing Reagents Employed in Gravimetric Methods

* Organic precipitating agents

A number of organic reagents have been developed for the gravimetric analysis of

inorganic species. In general, these reagents tend to be more selective in their reactions

than many of the inorganic reagents listed in Table 3.1.

Two types of organic reagents are encountered. One forms slightly soluble

nonionic complexes called coordination compounds. The other forms products in

which the bonding between the inorganic species and the reagent is largely ionic.

Organic reagents which yield sparingly soluble coordination compounds typically

contain at least two functional groups, each of which is capable of bonding with the

cation by donation of a pair of electrons. The functional groups are located in the

molecule in such a way that a five- or six-membered ring results from reaction.

50

Coordination compounds which form complexes of this type are called chelating

agents; their products with a cation are termed chelates.

Neutral coordination compounds are relatively nonpolar; as a consequence, their

solubilities are low in water but high in organic liquids. Chelates usually possess low

densities and are often intensely colored. Because they are not wetted by water,

coordination compounds are readily freed of moisture at low temperatures. At the

same time, however, their hydrophobic nature endows these precipitates with the

annoying tendency to creep up the sides of the filtering medium during the washing

operation; physical loss of solid may result unless care is taken. Three examples of

coordination reagents are considered here.

8-Hydroxyquinoline. Approximately two dozen cations form sparingly soluble

coordination compounds with 8-hydroxyquinoline, which is also known as oxine.

Typical of these is the product with magnesium:

The solubilities of metal oxinates vary widely from cation to cation and, moreover, are

pH dependent because proton formation always accompanies the chelation reaction.

Therefore, by control of pH, a considerable degree of selectivity can be imparted to 8-

hydroxyquinoline.

α-Nitroso-β-naphthol. This was one of the first selective organic reagents; its

structure is:

51

The reagent reacts with cobalt(II) to give a neutral cobalt(III) chelate having the

structure CoA3, where A- is the conjugate base of the reagent. Note that formation of

the product involves both oxidation and precipitation of the cobalt by the reagent; the

precipitate is contaminated by reduction products of the reagent as a consequence.

Therefore, it is common practice to ignite the chelate in oxygen to produce Co3O4;

alternatively, the ignition is performed in a hydrogen atmosphere to produce the

element as the weighed form.

The most important application of α-nitroso-β-naphthol has been for the

determination of cobalt in the presence of nickel. Other ions that precipitate with the

reagent include bismuth(III), chromium(III), mercury(II), tin(IV), titanium(III),

tungsten(VI), uranium(VI), and vanadium(V).

Dimethylglyoxime. An organic precipitating agent of unparalleled specificity is

dimethylglyoxime.

Its coordination compound with palladium is the only one that is sparingly soluble in

acid solution. Similarly, only the nickel compound precipitates from a weakly alkaline

environment. Nickel dimethylglyoxime is bright red and has the structure:

52

Sodium Tetraphenylboron. Sodium tetraphenylboron, (C6H5)4B-Na+, is an important

example of organic precipitating reagent that form saltlike precipitates. In cold mineral

acid solutions, it is a near-specific precipitating agent for potassium ion and for

ammonium ion. The precipitates are stoichiometric, corresponding to the potassium or

the ammonium salt, as the case may be; they are amenable to vacuum filtration and

can be brought to constant weight at 105 to 120oC. Only mercury(II), rubidium, and

cesium interfere and must be removed by prior treatment.

Benzidine. Another salt-forming reagent is benzidine.

Benzidine precipitates sulfate from a slightly acidic medium as C12H12N2.H2SO4.

The solubility of this precipitate increases rapidly with temperature and also with the

acidity of the environment; both variable must be carefully controlled. Instead of being

weighed as a gravimetric precipitate, benzidine sulfate may be titrated with a standard

solution of sodium hydroxide. Yet another method for completion of the analysis calls

for titration of the benzidine with a standard solution of permanganate. The methods

succeed in the presence of copper, cobalt, nickel, zinc, manganese(II), iron(II),

chromium(III), and aluminum ions. Benzidine is well suited to the rapid, routine

analysis of sulfate.

54

4.1 Introduction

The reactions used in analytical chemistry never result in complete conversion of

reactants to products. Instead, they proceed to a state of chemical equilibrium in

which the ratio of concentrations of reactants and products is constant. Equilibrium-

constant expressions are algebraic equations that describe the concentration

relationships existing among reactants and products at equilibrium. Among other

things, equilibrium-constant expressions permit calculation of the error in an analysis

resulting from the quantity of unreacted analyte that remains when equilibrium has

been reached.

4.2 Equilibrium Constant Expressions

The influence of concentration (or pressure if the species are gases) on the position

of a chemical equilibrium is conveniently described in quantitative terms by means of

an equilibrium-constant expression. Such expressions are derived from

thermodynamics. They are important because they permit the chemist to predict the

direction and completeness of a chemical reaction. An equilibrium-constant

expression, however, yields no information concerning the rate at which equilibrium is

approached. In fact, we sometimes encounter reactions that have highly favorable

equilibrium constants but are of little analytical use because their rates are low. This

limitation can often be overcome by the use of a catalyst, which speeds the attainment

of equilibrium without changing its position.

For the reaction:

mA + nB yC + zD

the equilibrium constant is:

[C]y[D]z

Kc = ـــــــــــــــــــــــ

[A]m[B]n

where Kc is the equilibrium constant, the letter c denote the concentrations in

mole/liter. The square-bracketed terms mean the molar concentration if the species is a

55

dissolved solute. And mean partial pressure in atmospheres if the species is a gas; in

fact, we often replace Kc with Kp in this case. Kc is large if the products more than

reactants, and small if the products less than reactants, and equal to 1 if the

concentrations of the products and reactants are equal at the equilibrium state.

Example 1

Write the equilibrium constant for the following reactions:

1- 2NO(g) + Cl2(g) 2NOCl(g)

2- Na2CO3(s) + BaSO4(s) Na2SO4(s) + BaCO3(s)

[NOCl(g)]2

1- Kp = ــــــــــــــــــــــــــــــــ

[NO(g)]2 [Cl2(g)]

[Na2SO4(s)][BaCO3(s)]

2- Kc = ــــــــــــــــــــــــــــــــــــــــــــــ

[Na2CO3(s)][BaSO4(s)]

Example 2

In the thermal cracking of HI at 321.4oC, the premier concentration of the substance

was 2.08 M, and in the equilibrium state the concentration was 1.68M. Calculate Kc.

2HI(g) H2(g) + I2(g)

at the beginning of the reaction the concentration of HI was 2.08 M, then become 1.68

M in the equilibrium state, so the disappeared quantity is:

2.08 – 1.68 = 0.40 M

If we go back to the balanced equation we find that the disappearance of 2 mole from

HI gives 1 mole from H2 and 1 mole from I2, then the concentration of H2 and I2 at the

equilibrium 0.40/2 = 0.2, so:

]2][I2[H

Kc = ــــــــــــــــــ

[HI]2

56

2HI(g) H2(g) + I2(g)

the concentration at the beginning 2.08 0 0

the concentration at the equilibrium 1.68 0.20 0.20

(0.20)(0.20)

Kc = 1.4 = 0.014 = ـــــــــــــــــــــــــ x 10-2

(1.68)2

4.3 Dissociation of Water

Water is weak electrolyte, and the aqueous solutions contain small concentrations

of hydronium H3O+, and hydroxide OH- ions as a consequence of the dissociation

reaction:

2H2O(l) H3O+

(aq) + OH-(aq)

or H2O(l) H+(aq) + OH-

(aq)

The equilibrium constant of this reaction is:

[H+][OH-]

Keq = ـــــــــــــــــــــــــــ

[H2O]

The concentration of water in dilute aqueous solutions is enormous when compared

with the concentration of hydrogen and hydroxide ions. As a consequence, [H2O] in

the equation can be taken as constant:

Keq [H2O] = Kw = [H+][OH-]

Where Kw is the ion-product constant for water.

At 25oC, the ion-product constant for water is 1.00 × 10-14, so:

[H3O+] = 1.0 × 10-7 M

[OH-] = 1.0 × 10-7 M

57

Example 3

If the concentration of OH- ion in an ammonia solution was 0.0025 M, calculate the

concentration of H+ ion.

Kw = [H+][OH-]

Kw

[H+] = ـــــــــــــــــ

[OH-]

1.0 × 10-14

[H+] = 4.0 = ــــــــــــــــــــــــ x 10-12 M

0.0025

Example 4

In a 1.0 × 10-4 M HCl solution, the concentration of hydrogen ions [H+] was 1.0 × 10-4,

calculate the hydroxide ions concentration. The ionization constant of water at 25oC is

1.0 × 10-14.

Kw = [H+][OH-]

1.0 x 10-14 = (1.0 x 10-4) [OH-]

1.0 x 10-14

[OH-] = 1.0 = ـــــــــــــــــــــ x 10-10 M

1.0 x 10-4

4.4 pH Concept

Because the concentrations of H+ and OH- ions in aqueous solutions are very

small, scientists invented the pH concept which is the negative logarithm of hydrogen

ion concentration in mole/liter. The negative logarithm gives positive number for the

pH.

]H

1[ pH = log

pH = - log [H+]

The pH is a measure of hydrogen ion concentration, so we can identify the solutions of

acids and bases at 25oC by knowing the pH:

58

acidic solutions: [H+] >1.0 ×10-7 M, pH < 7.00

basic solutions: [H+] <1.0 ×10-7 M, pH > 7.00

neutral solutions: [H+] =1.0 ×10-7 M, pH = 7.00

The pOH is the negative logarithm of hydroxide ion concentration in the solution:

pOH = - log [OH-]

[H+][OH-] = Kw = 1.0 × 10-14

If we take the negative logarithm for both sides of the equation, we get:

- (log [H+] + log [OH-]) = - log (1.0 × 10-14)

- log [H+] – log [OH-] = 14.00

From the definition of pH and pOH we get:

pH + pOH = 14

In neutral solution pH + pOH = 7

Example 5

The concentration of the hydrogen ion H+ in a solution was 3.2 × 10-4, calculate the

pH of the solution.

pH = - log [H+]

= - log (3.2 x 10-4) = 3.49

Example 6

The pH of a solution of acidic rain was 4.82. Calculate the concentration of H+ ion for

this solution.

pH = - log [H+]

4.82 = - log [H+]

[H+] = 1.5 x 10-5 M

59

Example 7

Calculate the pH of a 0.10 M NaOH solution.

NaOH is a strong electrolyte, so the value of OH- comes from the strong electrolyte.

[OH-] = 0.10 M

Kw = [H+][OH-]

Kw

[H+] = ـــــــــــــــــ

[OH-]

1.0 x 10-14

[H+] = 1.0 = ــــــــــــــــــــــ x 10-13 M

0.10

pH = - log [H+]

= - log (1.0 x 10-13) = 13

Example 8

In a solution of NaOH, the concentration of OH- ion was 2.9 × 10-4 M. Calculate the

pH of the solution.

pOH = - log [OH-]

= - log (2.9 x 10-4)

= 3.54

pH + pOH = 14.00

pH = 14 – pOH

= 14 – 3.54 = 10.46

60

4.5 Dissociation of Weak Acids and Bases

When a weak acid or a weak base is dissolved in water, partial dissociation occurs.

Thus if we assume that a weak acid HA dissociates according to the equation:

HA(aq) + H2O(aq) H3O+

(aq) + A-(aq)

or in a simpler way:

HA(aq) H+(aq) + A-

(aq)

the equilibrium constant is:

[H+][A-]

Ka = ــــــــــــــــــــ

[HA]

where Ka is the acid-dissociation constant. The stronger the acid the higher the value

of Ka. Only weak acids have dissociation constant. Because the values of dissociation

constants are very small, we use its logarithm, we call it mathematically pKa.

pKa = - log Ka

Table 4-1 contains values of Ka for some weak acids.

Table 4-1 Values of Ka for some Weak Acids

Acid Value aK

Nitrous acid -2+ NO +H 2HNO 4-4.5 x 10

c acidFormi -+ HCOO+ H HCOOH 4-1.7 x 10

Benzoic acid -COO5H6+ C +H COOH 5H6C 5-6.5 x 10

Acetic acid -COO3+ CH +H COOH3CH 5-1.8 x 10

Hydrocyanic acid -+ CN +H HCN 10-6.2 x 10

Weak polyprotic acids dissociate on steps, and each step has its own dissociation

constant. Carbonic acid H2CO3 for example dissociates on two steps:

[H+][HCO3-]

H2CO3 H+ + HCO3- Ka1 = ــــــــــــــــــــــــ

[H2CO3]

61

[H+][CO32-]

HCO3- H+ + CO3

2- Ka2 = ــــــــــــــــــــــــ

[HCO3-]

In an analogous way bases dissociate, and if a weak base dissolved in aqueous

solution, it dissociates for small degree then reaches equilibrium:

B + H2O BH+ + OH-

The dissociation constant is:

]-][OH+[BH

Kb = ـــــــــــــــــــــــــــ

[B]

Note that a concentration term for water ([H2O]) does not appear in the denominator

of either equation because it small and can be considered constant.

Example 9

Calculate Ka for acetic acid if premier concentration was 0.1 mol/L and the percentage

of dissociation is 1.34%.

percentage of dissociation is 1.34, this means:

in every 100 mole, 1.34 mole dissociate

in every 0.1 mole X mole dissociate

1.34 x 0.1

X = ــــــــــــــــــــــــــــ = 0.00134

100

CH3COOH CH3COO- + H+

concentration at the beginning 0.10 0 0

concentration at the equilibrium 0.1- 0.00134 0.00134 0.00134

[CH3COO-][H+]

Ka = ــــــــــــــــــــــــــــــــــــ

[CH3COOH]

62

(0.00134)(0.00134)

ـــــــــــــــــــــــــــــــــــــــــــــــ =

0.10 – 0.00134

= 1.82 x 10-5

Example 10

If the Ka for acetic acid is 1.8 × 10-5, calculate the molar concentration for hydrogen

ions [H+], and the dissociation percentage for 0.50 M acetic acid.

suppose x is the number of dissociated moles of acid per liter, so:

x = number of H+ and CH3COO- moles produced per each liter, and (0.50 – x) =

number of undissociated CH3COOH moles per liter.

CH3COOH CH3COO- + H+

concentration at the beginning 0.50 0 0

concentration at the equilibrium 0.50 - x x x

[CH3COO-][H+]

Ka = 1.8 x 10-5 = ـــــــــــــــــــــــــــــــــــ

[CH3COOH]

(x)(x)

ـــــــــــــــــــــــ =

0.50 – x

The dissociation of weak acetic acid is very small, so the concentration of x is very

small in contrast with the concentration of undissociated acid. In this example the

value (0.50 – x) probably is very near to 0.50, so we can use a simple way and avoid

using the quadratic equation:

1.8 ×10-5 ≈ 50.0

))(( xx

x2 ≈ 0.90 ×10-5 = 9.0 ×10-6

63

x ≈ 6100.9

x ≈ 3.0 ×10-3 M = [H+]

it is also equal to the number of dissociated moles of acetic acid per liter.

number of dissociated moles

percentage of dissociated acid = ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ × 100

number of moles at the start

3.0 × 10-3

ـــــــــــــــــــــــــــــــ = = 0.60%

0.50

4.6 Complex Formation Constant

An analytically important class of reactions involves the formation of soluble

complex ion. Example is:

[Fe(SCN)2+]

Fe3+ + SCN- Fe(SCN)2+ Kf ـــــــــــــــــــــــــــــــ =

[Fe3+][SCN-]

where Kf is called the formation constant for the complex.

4.7 Oxidation-Reduction Equilibrium Constant

Equilibrium constant for oxidation-reduction reactions can be formulated in the

usual way. For example:

6Fe2+ + Cr2O72- + 14H3O

+ 6Fe3+ + 2Cr3+ + 21H2O

[Fe3+]6[Cr3+]2

K = ـــــــــــــــــــــــــــــــــــــــــــــــــ

[Fe2+]6[Cr2O72-][H3O

+]14

The concentration of water does not appear in the equilibrium since it is constant.

64

4.8 Le Chatelier's Principle

There is a general rule that helps us to predict the direction in which an

equilibrium reaction will move when a change in concentration, pressure, volume, or

temperature occurs. The rule, known as Le Chatelier's principle, state that if an

external stress is applied to a system at equilibrium, the system adjusts in such a way

that the stress is partially offset as the system reaches a new equilibrium position. The

word stress here means a change in concentration, pressure, volume, or temperature

that removes a system from the equilibrium state.

4.9 Factors Affect Chemical Equilibrium

1- Effect of pressure

Changes in pressure ordinary do not affect the concentrations of reacting species

in condensed phases (say, in an aqueous solution) because liquids and solids are

virtually incompressible. On the other hand, concentrations of gases are greatly

affected by changes in pressure.

Suppose that the equilibrium system,

N2O4 (g) 2NO2(g)

is in a cylinder fitted with a movable piston. What happens if we increase the pressure

on the gases by pushing down on the piston at constant temperature? Since the volume

decreases, the concentration of both NO2 and N2O4 increase. Because the

concentration of NO2 is squared in the equilibrium constant expression, the increase in

pressure increases the numerator more than the denominator. The system is no longer

at equilibrium.

2- Effect of temperature

The values of equilibrium are measured at certain temperature. Any change in

temperature will change the equilibrium constant. For example, the formation of NO2

from N2O4 is an endothermic process, and the reverse reaction is exothermic:

65

N2O4 (g) → 2NO2(g)

At equilibrium the net heat effect is zero because there is no net reaction. If the system

is heated at constant volume more N2O4 will dissociate to NO2, and that shift the

reaction to the right. In opposite, if the temperature is decreased, the reaction will shift

to the left.

3- Effect of concentration

If we consider the equilibrium:

CH3COOH + H2O H3O+ + CH3COO-

Any increase in H3O+ or CH3COO- concentration shifts the equilibrium to the left, and

any increase in CH3COOH concentration shifts the equilibrium to the right, but K

remains constant.

4- Effect of catalyst

In an equilibrium system, a catalyst increases the speed of both forward and

reverse reactions to the same extent. A catalyst does not change the relative amounts

present at equilibrium; the value of the equilibrium constant is not changed. The

catalyst dose change the time required for reaching the equilibrium. Reactions that

require days or weeks to come to equilibrium may reach it in a matter of minutes in

the presence of a catalyst.

4.10 Buffer Solutions

A solution that contains a weak acid plus a salt of that acid, or a weak base plus a

salt of that base, has the ability to react with both strong acids and strong bases. Such a

system is referred to as a buffer solution, because small additions of either strong acids

or strong bases produce little change in the pH.

A simple buffer solution can be prepared by adding comparable molar amounts of

acetic acid (CH3COOH) and its salt sodium acetate (CH3COONa) to water. This has

66

the ability to neutralize either added acid or added base. Sodium acetate, a strong

electrolyte, dissociate completely in water:

CH3COONa(s) → CH3COO-(aq) + Na+

(aq)

If an acid is added, the H+ ions will be consumed by the conjugate base in the

buffer, CH3COO-, according to the equation:

CH3COO-(aq) + H+

(aq) → CH3COOH(aq)

If a base is added to the buffer system, the OH- ions will be neutralized by the acid

in the buffer:

CH3COOH(aq) + OH-(aq) → CH3COO-

(aq) + H2O(l)

In general, a buffer system can be represented as salt-acid or conjugate base-acid.

Thus the sodium acetate-acetic acid buffer system can be written as

CH3COONa/CH3COOH or simply CH3COO-/CH3COOH.

4.11 Calculation of the pH of Buffer Solutions

A general buffer equation can be derived by considering the following reactions

for a weak acid, HA, and the salt of its conjugate weak base, NaA.

[H3O+][A-]

HA + H2O H3O+ + A- Ka = 1…..……… ـــــــــــــــــــــــــــ)

[HA]

[OH-][ HA] Kw

A- + H2O OH- + HA Kb = 2…… ــــــــــ = ـــــــــــــــــــــــــــ)

[A-] Ka

To find the pH of a solution containing both an acid, HA, and its conjugate base, NaA,

we need to express the equilibrium concentrations of HA and NaA in terms of their

analytical concentrations. CHA and CNaA. An examination of the two equilibria reveals

that the first reaction decreases the concentration o f HA by an amount equal to

67

[H3O+], whereas the second increases the HA concentration by an amount equal to

[OH-]. Thus, the species concentration of HA is related to its analytical concentration

by the equation:

[HA] = CHA – [H3O+] + [OH-] …….3)

Similarly, the first equilibrium will increase the concentration of A- by an amount

equal to [H3O+], and the second will decrease this concentration by the amount [OH-].

Thus, the equilibrium concentration is given by a second equation similar to Equation

3. [A-] = CNaA + [H3O

+] - [OH-] …….4)

Because of the inverse relationship between [H3O+] and [OH-], it is always

possible to eliminate one or the other from Equations 3 and 4. Moreover, the

difference in concentration between these two species is usually so small relative to

the molar concentrations of acid and conjugated base that Equations 3 and 4 simplify

to:

[HA] ≈ CHA ………5)

[A-] ≈ CNaA ………6)

Substitution of Equations 5 and 6 into the dissociation-constant expression and

rearrangement yields:

CHA

[H3O+] = Ka 7..………… ــــــــــــــــ)

CNaA

The assumption leading to Equations 5 and 6 sometimes breaks down with acids or

bases that have dissociation constants greater than about 10-3 or when the molar

concentration of either the acid or its conjugate base (or both) is very small. In these

circumstances, either [OH-] or [H3O+] must be retained in Equations 3 and 4,

depending on whether the solution is acidic or basic. In any case, Equations 5 and 6

68

should always be used initially. The provisional values for [ H3O+] and [OH-] can then

be used to test the assumptions.

* The Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation, which is used to calculate the pH of buffer

solutions, is frequently encountered in the biological literature and biochemical texts.

It is obtained by expressing each term in Equation 7 in the form of its negative

logarithm and inverting the concentration ratio to keep all signs positive:

CNaA

-log [H3O+] = - log Ka + log ــــــــــــــــ

CHA

Therefore,

CNaA

pH = pKa + log 8.…… ـــــــــــــــ)

CHA

Example 11

What is the pH of a solution that is 0.400 M in formic acid and 1.00 M in sodium

formate?

The pH of this solution will be affected by the Kw of formic acid and the Kb of

formate ion.

HCOOH + H2O H3O+ + HCOO- Ka = 1.80 × 10- 4

Kw

HCOO- + H2O HCOOH + OH- Kb = 11-10 × 5.56 = ــــــــــ

Ka

Since the Ka for formic acid is orders of magnitude larger than the Kb for formate, the

solution will be acidic and Ka will determine the H3O+ concentration. We can thus

write:

69

[H3O+][HCOO-]

Ka = 4-10 × 1.80 = ـــــــــــــــــــــــــــــــــ

[HCOOH]

[HCOO-] ≈ CHCOO- = 1.00 M

[HCOOH] ≈ CHCOOH = 0.400 M

Substitution into Equation 7 gives, with rearrangement:

0.400

[H3O M 5-10 × 7.20 = ـــــــــــــــ × 10-4 × 1.8 = [+

1.00

Note that the assumption that [H3O+] << CHCOOH and that [H3O

+] << CHCOO- is valid.

Thus:

pH = - log (7.20 × 10-5) = 4.14

4.12 Buffer Capacity

The buffer capacity of a solution is defined as the number of equivalents of strong

acid or strong base needed to cause 1.00 liter of the buffer to undergo a 1.00-unit

change in pH. The buffer capacity is dependent upon the concentration of the

conjugated acid-base pair. It is also dependent upon their concentration ratio, and

reaches maximum when this ratio is unity.

71

5.1 Introduction

A quantitative analysis based upon the measurement of volume is called a

volumetric or titrimetric method. Volumetric methods are much widely used than

gravimetric methods because they are usually more rapid and convenient; in addition,

they are often as accurate.

5.2 Definition of Some Terms

Titration is the process by which the quantity of analyte in a solution is determined

from the amount of a standard reagent it consumes. Ordinarily, a titration is performed

by carefully adding the reagent of known concentration until reaction with the analyte

is judged to be complete; the volume of standard reagent is then measured.

Occasionally, it is convenient or necessary to add an excess of the reagent and then

determine the excess by back-titration with a second reagent of known concentration.

The reagent of exactly known concentration that is used in a titration is called a

standard solution. The accuracy with which its concentration is known sets a define

limit upon the accuracy of the method; for this reason, much care is taken in the

preparation of standard solutions. The concentration of a standard solution is

established either directly or indirectly:

1- by dissolving a carefully weighed quantity of the pure reagent and dilution to an

exactly known volume.

2- by titrating a solution containing a weighed quantity of a pure compound with the

reagent solution.

In either method, a highly purified chemical compound called a primary standard

is required as the reference material. The process whereby the concentration of a

standard solution is determined by titration of a primary standard is called

standardization.

The goal of every titration is the addition of standard solution in an amount that is

chemically equivalent to the substance with which it reacts. This condition is achieved

at the equivalence point. For example, the equivalence point in the titration of sodium

72

chloride with silver nitrate is attained when exactly one formula weight of silver ion

has been introduced for each formula weight of chloride ion present in the sample. In

the titration of sulfuric acid with sodium hydroxide, the equivalence point occurs when

two formula weights of the latter have been introduced for each formula weight of the

former.

The equivalence point in a titration is a theoretical concept; in actual fact, its

position can be estimated only by observing physical changes associated with

equivalence. These changes manifest themselves at the end point of the titration. It is

to be hoped that any volume difference between the end point and the equivalence

point will be small. Differences do exist, however, owing to inadequacies in the

physical changes and our ability to observe them; a titration error is the result.

A common method of end-point detection in volumetric analysis involves the use

of a supplementary chemical compound that exhibits a change in color as a result of

concentration changes occurring near the equivalence point. Such a substance is called

an indicator.

5.3 Reactions and Reagents Used in Volumetric Analysis

It is convenient to classify volumetric methods according to four reaction types,

specifically, precipitation, complex formation, neutralization (acid-base), and

oxidation-reduction. Each reaction type is unique in such matters as nature of

equilibria involved; the indicators, reagents, and primary standard available; and the

definition of equivalent weight.

* Primary standards

The accuracy of a volumetric analysis is critically dependent upon the primary

standard used to establish, directly or indirectly, the concentration of the standard

solution. Important requirements for a substance to serve as a good primary standard

include the following:

73

1- Highest purity. Moreover, established methods should be available for confirming

its purity.

2- Stability. It should not be attacked by constituents of the atmosphere.

3- Absence of hydrate water. If the substance were hygroscopic or efflorescent,

drying and weighing would be difficult.

4- Ready availability at reasonable cost.

5- Reasonably high equivalent weight. The weight of a compound required to

standardize or prepare a solution of a given concentration increases directly with its

equivalent weight. Since the relative error in weighing decreases with increasing

weight, a high equivalent weight will tend to minimize weighing errors.

Few substances meet or even approach these requirements. As a result, the

number of primary-standard substances available to the chemist is limited.

In some instances, it is necessary to use less pure substances in lieu of a primary

standard. The assay (that is, the percent purity) of such a secondary standard must be

established by careful analysis.

5.4 Standard Solution

An ideal standard solution for titrimetric analysis would have the following

properties:

1- Its concentration should remain constant for months or years after preparation to

eliminate the need for restandardization.

2- Its reaction with the analyte should be rapid in order that the waiting period after

each addition of reagent does not become excessive.

3- The reaction between the reagent and the analyte should be reasonably complete.

4- The reaction of the reagent with the analyte must be such that it can be described by

a balanced chemical equation; otherwise, the weight of the analyte cannot be

calculated directly from the volumetric data. This requirement implies the absence of

side reactions between the reagent and the unknown or with other constituents of the

solution.

74

5- A method must exist for detecting the equivalence point between the reagent and

the analyte; that is, a satisfactory end point is required.

Few volumetric reagents currently in use meet all of these requirements perfectly.

5.5 End Point in Volumetric Methods

End points are based upon a physical property which changes in a characteristic

way at or near the equivalence point in the titration. The most common end point

involves a color change due to the reagent, the analyte, or an indicator substance.

Other physical properties, such as electrical potential, conductivity, temperature, and

refractive index, have also been employed to locate the equivalence point in titration.

5.6 Theory of Neutralization Titration for Simple Systems

End-point detection in a neutralization titration is ordinary based upon the sharp

change in pH that occurs near the equivalence point. The pH range with which such a

change occurs varies from titration to titration and is determined both by the nature

and the concentration of the analyte as well as the titrant. The selection of an

appropriate indicator and the estimation of the titration error require knowledge of the

pH changes which occur throughout the titration. Thus, we need to know how

neutralization titration curves are derived.

5.7 Standard Solutions for Neutralization Titrations

The standard solutions employed for neutralization titrations are always strong

acids or strong bases because these react more completely than their weaker

counterparts, and they therefore provide sharper end point. Standard solutions of acids

are prepared by diluting concentrated acids.

Standard solutions of bases are ordinarily prepared from solid bases.

5.8 Acid-Base Indicators

An acid-base indicator is a weak organic acids or a weak organic base whose

undissociated form differ in color from its conjugate base or its conjugate acid form.

75

For example, the behavior of an acid-type indicator, HIn, is described by the

equilibrium:

HIn + H2O In- + H3O+

acid color base color

Here, internal structure changes accompany dissociation and cause the color change.

The equilibrium for a base-type indicator, In, is:

In + H2O InH+ + OH-

base color acid color

The equilibrium-constant expression for the dissociation of acid-type indicator

takes the form:

[H3O+][In-]

Ka = ــــــــــــــــــــــــــ

[HIn]

[HIn]

[H3O+]Ka = ـــــــــــــــــــ

[HI-]

We then see that the hydronium ion concentration determines the ratio of the acid to

the conjugate base from the indicator, which in turn controls the color of the solution.

The human eye is not very sensitive to color differences in a solution containing a

mixture of HIn and In-, particularly when the ratio [HIn]/[In-] is greater than about 10

or smaller than about 0.1. Consequently, the color change detected by an average

observer occurs within a limited range of concentration ratios from about 10 to about

0.1. At greater or smaller ratios, the color appears essentially constant to the eye and is

independent of the ratio. As a result, we can write that the average indicator, HIn,

exhibits its pure acid color when:

[HIn] 10

ــــــــــــــــــ ـــــــــــــ ≤

[In-] 1

and its base color when:

76

[HIn] 1

ــــــــــــــــــ ـــــــــــــ ≥

[In-] 10

The color appears to be intermediate for ratios between these two values.

5.9 The Common Acid-Base Indicators

The list of acid-base indicators is large and includes a number of organic

compounds. An indicator covering almost any desired pH range can ordinarily be

found. A few common indicators are given in Table 5-1.

Table 5-1 Some Important Acid-Base Indicators

Common Name Transition

Range, pH

Color Change

Acid Base

Indicator

Type

Thymol blue 1.2-2.8

8.0-9.6

red

yellow

yellow

blue 1

Methyl yellow 2.9-4.0 red yellow 2

Methyl orange 3.1-4.4 red orange 2

Bromocresol green 3.8-5.4 yellow blue 1

Methyl red 4.2-6.3 red yellow 2

Bromocresol purple 5.2-6.8 yellow purple 1

Bromothymol blue 6.2-7.6 yellow blue 1

Phenol red 6.8-8.4 yellow red 1

Cresol purple 7.6-9.2 yellow purple 1

Phenolphthalein 8.3-10.0 colorless red 1

Thymolphthalein 9.3-10.5 colorless blue 1

Alizarin yellow GG 10-12 colorless yellow 2

(1) Acid type HIn + H2O H3O+ + In-

(2) Base type In + H2O InH+ + OH-

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5.10 Titration of a Strong Acid with a Strong Base

Strong acids and strong bases are completely ionized in their solutions, examples

HCl and NaOH.

HCl + NaOH NaCl + H2O

At the equivalence point, the hydronium and hydroxide ions are present in equal

concentrations. The solution containing NaCl salt does not undergo hydrolysis, and the

pH of the solution is 7.00.

If the values of pH are plotted against the volumes of NaOH added, we obtain

curve as shown in Figure 5-1, called titration curve for a strong acid with strong base.

* The effect of concentration

The effect of reagent and analyte concentrations on the neutralization titration

curves for strong acids are shown by the plots in Figure 5-1. Note that with 0.1 M

NaOH as the titrant, the change in pH in the equivalence-point region is large. With

0.001 M NaOH, the change is markedly less but still pronounced.

* Choosing an indicator

Figure 5-1 shows that the selection of an indicator is not critical when the reagent

concentration is approximately 0.1 M.

Here, the volume differences in titrations with the three indicators shown are of the

same magnitude as the uncertainties associated with reading the buret; therefore, they

are negligible. Note, however, that bromocresol green is unsuited for a titration

involving the 0.001 M reagent because the color change occurs over a 5-mL range

well before the equivalence point. The use of phenolphthalein is subject to similar

objections. Of the three indicators, then, only bromothymol blue provides a

satisfactory end point with a minimal systematic error in the titration of the more

dilute solution.

Titration curves for strong bases are derived in an analogous way to those for

strong acids. The solution is neutral at the equivalence point.

78

Figure 5-1 Titration Curves for HCl with NaOH. Curve A: 50.00 mL of 0.0500 M HCl with

0.1000 M NaOH. Curve B: 50.00 mL of 0.000500 M HCl with 0.001000 M NaOH.

5.11 Titration of a weak Acid with a Strong Base

Weak acid, like acetic acid, does not ionize in the solution completely. The

equation for the reaction is:

CH3COOH + NaOH CH3COONa + H2O

At the beginning, the solution contains only a weak acid or weak base, and the pH

is calculated from the concentration of the solute and its dissociation constant.

At the equivalence point, the solution contains only the salt, and the pH is

calculated from the concentration of this product. The solution containing CH3COONa

salt, undergoes hydrolysis as shown in the equation:

CH3COONa + H2O CH3COOH + NaOH

which increases the hydroxide ions concentration over the concentration of hydrogen

ions, because NaOH ionizes completely and the CH3COOH ionizes partially. The pH

of the solution is 8.72. The titration curve is shown in Figure 5-2.

79

* The effect of concentration

The initial pH values are higher and the equivalence-point pH is lower for the

more dilute solution (curve B).

* Choosing an indicator

Figure 5-2 shows that the choice of indicator is more limited for the titration of a

weak acid than for the titration of a strong acid. The bromocresol green is totally

unsuited for titration of 0.1000 M acetic acid. Bromothymol blue does not work either

because its full color change occurs over a range of titrant volume from about 47 mL

to 50 mL of 0.1000 M base. An indicator exhibiting a color change in the basic region,

such as phenolphthalein, however, should provide a sharp end point with a minimal

titration error.

Figure 5-2 Titration Curves for CH3COOH with NaOH. Curve A: 0.1000 M acid with of 0.1000

M base. Curve B: 0.001000 M acid with 0.001000 M base.

80

The end-point pH associated with the titration of 0.001000 M acetic acid (curve B)

is so small that a significant titration error is likely to be introduced regardless of

indicator. Use of an indicator with a transition range between that of phenolphthalein

and that of bromothymol blue in conjunction with a suitable color comparison

standard, however, makes it possible to establish the end point in this titration with

decent precision.

5.12 Titration of a Strong Acid with a Weak Base

The calculations needed to draw the titration curve for a weak base are analogous

to those for a weak acid.

Consider the titration of HCl, (a strong acid) and ammonium hydroxide (a weak

base). The equation for the reaction is:

HCl + NH4OH NH4Cl + H2O

The pH at the equivalence point, is less than 7.00 (5.28), due to the hydrolysis of

the ammonium chloride, and formation of HCl, which is more ionized than NH4OH.

NH4Cl + H2O NH4OH + HCl

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6.1 Introduction

Complex systems are defined as solutions made up of:

(1) two acids or two bases of different strength.

(2) an acid or a base that has two or more acidic or basic functional groups.

(3) an amphiprotic substance, which is capable of acting as both an acid and a base.

Polyfunctional acids and bases play important roles in many chemical and

biological systems. The human body contains a complicated system of buffers within

cells and within bodily fluids, such as human blood. The pH of human blood is

controlled to be within the range of 7.35 to 7.45 primarily by the carbonic

acid/bicarbonate buffer system.

6.2 Mixture of Strong and Weak Acids or Strong and weak Bases

It is possible to determine each of the components in a mixture containing a strong

acid and a weak acid (or a strong base and a weak base) provided that the

concentrations of the two are the same order of magnitude and that the dissociation

constant for the weak acid or base is somewhat less than about 10-4. To demonstrate

that this statement is true, let us show how a titration curve can be constructed for a

solution containing roughly equal concentrations of HCl and HA, where HA is a weak

acid with dissociation constant of 10-4.

When the amount of base added is equivalent to the amount of hydrochloric acid

originally present, the solution is identical in all respects to one prepared by dissolving

appropriate quantities of the weak acid and sodium chloride in a suitable volume of

water. The sodium chloride, however, has no effect on the pH (neglecting the

influence of increased ionic strength); thus, the reminder of the titration curve is

identical to that for a dilute solution of HA.

The shape of the curve for a mixture of weak and strong acids, and hence the

information obtainable from it, depends in large measure on the strength of the weak

acid. Figure 6-1 depicts the pH change that occurs during the titration of mixtures

containing hydrochloric acid and several weak acids.

83

Note that the rise in pH at the first equivalence point is small or essentially nonexistent

when the weak acid has a relatively large dissociation constant (curve A and B). For

titrations such as these, only the total number of millimoles of weak and strong acid

can be determined accurately. Conversely, when the weak acid has a very small

dissociation constant, only the strong acid content can be determined. For weak acids

of intermediate strength (Ka somewhat less than 10-4 but greater than 10-8), there are

usually two useful end points.

Figure 6-1 Curves for the Titration of Strong Acid/Weak Acid Mixtures with 0.1000 M NaOH.

Each titration is on 25.00 mL of a solution that is 0.1200 M in HCl and 0.0800 M in HA.

Determination of the amount of each component in a mixture that contains a strong

base and a weak base is also possible, subject to the constraints just described for the

strong acid/weak acid system. The computation of a curve for such a titration is

analogous to that for a mixture of acids.

84

6.3 Polyfunctional Acids and Bases

Several species are encountered in analytical chemistry that have two or more

acidic functional groups. Generally, the two groups differ in strength and, as a

consequence, exhibit two or more end points in a neutralization titration.

* The phosphoric acid system

Phosphoric acid is a typical polyfunctional acid. In aqueous solution, it undergoes

the following three dissociation reactions:

[H3O+][H2PO4

-]

H3PO4 + H2O H2PO4- + H3O

+ Ka1 = ــــــــــــــــــــــــــــــــ [H3PO4]

= 7.11 × 10-3

[H3O+][HPO4

2-]

H2PO4- + H2O HPO4

2- + H3O+ Ka2 = ــــــــــــــــــــــــــــــــــ

[H2PO4-]

= 6.32 × 10-8

[H3O+][PO4

3-]

HPO42- + H2O PO4

3- + H3O+ Ka3 = ــــــــــــــــــــــــــــــــ

[HPO42-]

= 4.5 × 10-13

With this acid, as with other polyprotic acids, Ka1 ˃ Ka2 ˃ Ka3.

When we add two adjacent stepwise equilibria, we multiply the two equilibrium

constant to obtain the equilibrium constant for the resulting overall reaction. Thus, for

the first two dissociation equilibria for H3PO4, we write:

[H3O+]2[HPO4

2-]

Ka1Ka2 = ــــــــــــــــــــــــــــــــــ

[H3PO4]

85

= 7.11 × 10-3 × 6.32 × 10-8 = 4.49 × 10-10

Similarly, for the reaction:

H3PO4 3H3O+ + PO4

3-

we may write:

[H3O+]3[PO4

3-]

Ka1Ka2Ka3 = ــــــــــــــــــــــــــــــــــ

[H3PO4]

= 7.11 × 10-3 × 6.32 × 10-8 × 4.5 × 10-13 = 2.0 × 10-22

* The carbon dioxide carbonic acid system

When carbon dioxide is dissolved in water, a dibasic acid system is formed by the

following reaction:

[H2CO3]

CO2(aq) + H2O H2CO3 Khyd = 3-10 × 2.8 = ـــــــــــــــــــــــ

[CO2(aq)]

[H3O+][HCO3

-]

H2CO3+ H2O H3O+ + HCO3

- K1 = 4-10 × 1.5 = ـــــــــــــــــــــــــــــــــ

[H2CO3]

[H3O+][CO3

2-]

HCO3- + H2O H3O

+ + CO32- K2= 11-10× 4.69 = ــــــــــــــــــــــــــــــــ

[HCO3-]

The first reaction describes the hydration of aqueous CO2 to form carbonic acid.

Note that the magnitude of Khyd indicates that the concentration of CO2(aq) is much

larger than the concentration of H2CO3 (that is, [H2CO3] is only about 0.3% that of

[CO2(aq)]. Thus, a more useful way of discussing the acidity of solutions of carbon

dioxide is to combine the first and the second equations to give:

[H3O+][HCO3

-]

CO2(aq) + 2H2O H3O+ +HCO3

- Ka1 = ــــــــــــــــــــــــــــــــــ [CO2(aq)]

86

= 2.8 × 10-3 × 1.5 × 10-4

= 4.2 × 10-7

HCO3- + H2O H3O

+ + CO32- Ka2 = 4.69 ×10-11

6.4 Titration Curves for Polyfunctional Acids

Compounds with two or more acid functional groups yield multiple end points in a

titration provided that the functional groups differ sufficiently in strength as acids.

Figure 6-2 shows the titration curve for diprotic acid H2A with dissociation

constant of Ka1 = 1.00 × 10-3 and Ka2 = 1.00 × 10-7.

We can calculate each equivalence point for reaction like this.

Figure 6-2 Titration of 20.00 mL of 0.1000 M H2A with 0.1000 M NaOH. For H2A, Ka1 = 1.00 ×

10-3 and Ka2 = 1.00 × 10-7. The method of pH calculation is shown for several points regions on

the titration curve.

88

7.1 Introduction

Precipitation titrimetry, which is based on reactions that yield compounds of

limited solubility, is one of the oldest analytical techniques, dating back to the mid-

1800s. Because of the slow rate of formation of most precipitates, however, there are

only a few precipitating agents that can be used in titrimetry. The most important

precipitating reagent is silver nitrate, which is used for the determination of the

halides, the halide-like anions (SCN-, CN-, CNO-), mercaptans, fatty acids, and several

divalent and trivalent inorganic anions. Titrimetric methods based on silver nitrate are

sometimes called argentometric methods.

7.2 Precipitation Titration Curves Involving Silver Ion

The most common method of determining the halide ion concentration of aqueous

solutions is titration with a standard solution of silver nitrate. The reaction product is

solid silver halide. A titration curve for this method usually consists of a plot pAg

versus the volume of silver nitrate added. To construct titration curves, three types of

calculations are required, each of which corresponds to a distinct stage in the reaction:

(1) preequivalence, (2) equivalence, and (3) postequivalence.

7.3 The Effect of Concentration on Titration Curves

The effect of reagent and analyte concentrations on titration curves was shown by

the two titration curves in Figure 7-1. With 0.1 M AgNO3 (Curve A), the change in

pAg in the equivalence-point region is large. With the 0.01 M reagent, the change is

markedly less but still pronounced. Thus, an indicator for Ag+ that produces a signal in

the 4.0 to 6.0 pAg range should give a minimum error for the stronger solution. For

the more dilute chloride solution, the change in pAg in the equivalence-point region

would be too small to be detected precisely with a visual indicator.

89

Figure 7-1 Titration Curve for A, 50.00 mL of 0.0500 M NaCl with 0.1000 M AgNO3, and B,

50.00 mL of 0.00500 M NaCl with 0.0100 M AgNO3.

7.4 Indicators for Argentometric Titrations

Three types of end points are encountered in titrations with silver nitrate: (1)

chemical, (2) potentiometric, and (3) amperometric. Potentiometric end points are

obtained by measuring the potential between a silver electrode and a reference

electrode whose potential is constant and independent of the added reagent. To obtain

an amperometric end point, the current generated between a pair of silver

microelectrodes in the solution of the analyte is measured and plotted as a function of

reagent volume.

The end point produced by a chemical indicator usually consists of a color change

or, occasionally, the appearance or disappearance of turbidity in the solution being

90

titrated. The requirements for an indicator for a precipitation titration are that (1) the

color change should occur over a limited range in p-function of the titrant or the

analyte and (2) the color change should take place within the steep portion of the

titration curve for the analyte.

7.5 Mohr’s Method

Mohr’s method for chloride determination depends on the titration reaction:

Ag+ + Cl- AgCl(s) white ppt.

When all Cl- ions react with Ag+ ions, the first extra drop of Ag+ will react with CrO42-

ions of the indicator forming red precipitate, and that is the end point of the reaction.

2Ag+ + CrO42- Ag2CrO4(s) red ppt.

Sodium chromate can serve as an indicator for argentometric determination of

chloride, bromide, and cyanide ions by reacting with silver ion to form a brick-red

silver chromate (Ag2CrO4) precipitate in the equivalence-point region.

Silver chromate do not precipitate at the beginning, although its solubility

constant is smaller than that of silver chloride, and that is because the required amount

of silver ions to precipitate the chloride ions is 1.82×10-7 mol/L, which is much less

than the amount of silver ions required for precipitation of chromate ions (3.3×10-5

mol/L). Consequently, chromates precipitate after precipitation of chloride ions.

The concentration of chromate ions has an important role in determining the end

point. From the calibration curve for silver nitrate with chloride we see that the

equivalence point located at pAg = 4.87, that is, the concentration of silver ion is equal

to 1.35×10-5 mol/L, accordingly, the chromate ions concentration required to initiate

the precipitation of silver chromate is equal to 6.0×10-3 mol/L. In principle, this

amount of chromate ions is required to start the formation of the red brick precipitate

of silver chromate. This is by simply adding a slight access of the silver solution after

the equivalence point. But, since the chromate gives a bright yellow color at this

concentration, which makes it impossible to see the red color of silver chromate easily,

91

it is preferable and necessary to add a concentration less than 6.0×10-3 mol/L. This, of

course, requires a greater concentration of the silver needed to form the red brick

precipitate. This results in adding an excess amount of silver nitrate, especially in

dilute solutions, so that the precipitate can be seen clearly. In both cases, there will be

some error in the calibration. To overcome this difficulty, the volume taken must be

corrected by titrating a volume approximately equal to the volume of the total solution

at the equivalence point containing calcium carbonate and equivalent amount of the

indicator then subtracting the volume of silver nitrate needed to form silver chromate

precipitate, from the volume of silver reacted with the chloride.

The Mohr titration must be carried out at pH of 7 to 10 because chromate ion is the

conjugate base of the weak chromic acid, react with hydrogen:

CrO42- + H+ HCrO4

-

2HCrO4- H2O + Cr2O7

-2

In a strong alkaline solution, the silver ions precipitate as silver oxide:

2Ag+ + 2OH- 2AgOH Ag2O + H2O

Normally, a suitable pH is achieved by saturating the analyte solution with sodium

hydrogen carbonate.

7.6 Fajan’s Method

Fagan’s method also called adsorption indicator, this indicator is an organic

compound that tends to be adsorbed onto the surface of the solid in a precipitation

titration. Ideally, the adsorption (or desorption) occurs near the equivalence point and

results not only in a color change but also in a transfer of color from the solution to the

solid (or the reverse).

Fluorescein is a typical adsorption indicator that is useful for the titration of

chloride ion with silver nitrate. In aqueous solution, fluorescein partially dissociates

into hydronium ions and negatively charged fluoresceinate ions that are yellow-green.

92

The fluoresceinate ion forms an intensely red silver salt. Whenever this dye is used as

an indicator, however, its concentration is never large enough to precipitate as silver

fluoresceinate.

In the early stage of the titration of chloride ion with silver nitrate, the colloidal

silver chloride particles are negatively charged because of adsorption of excess

chloride ions (Figure 7.2). The dye anions are repelled from this surface by

electrostatic repulsion and impart a yellow-green color to the solution.

Figure 7-2 Fajan’s Method for the Titration of Chloride

Beyond the equivalence point, however, the silver chloride particles strongly adsorb

silver ions and thereby have a positive charge. Fluoresceinate anions are now attracted

into the counter-ion layer that surrounds each colloidal silver chloride particle. The net

result is the appearance of the red color of silver fluoresceinate in the surface layer of

the solution surrounding the solid. It is important to emphasize that the color change is

an adsorption (not a precipitation) process, because the solubility product of silver

fluoresceinate is never exceeded. The adsorption is reversible, the dye being desorbed

on back-titration with chloride ion.

Titrations involving adsorption indicators are rapid, accurate, and reliable, but their

application is limited to the relatively few precipitation reactions in which a colloidal

precipitate is formed rapidly.

93

7.7 Volhard’s Method

The most important application of the Volhard’s method is the indirect

determination of halide ions. A measured excess of standard silver nitrate solution is

added to the sample, and the excess silver is determined by back-titration with a

standard thiocyanate solution.

At the beginning the following reaction takes place:

Ag+ + Cl- AgCl(s)↓

Then the excess Ag+ ions are determined by titration with standard solution of KSCN.

Ag+ + SCN- AgSCN(s)

When all the Ag+ ions are reacted SCN- ions, the first excess drop of SCN- will react

with iron(III) ions of the indicator forming red complex indicating the end point of the

reaction.

[Fe(SCN)]2+

Fe3+ + SCN- [Fe(SCN)]2+ Kf = 1.05 × 103 = ــــــــــــــــــــــــــــ red [Fe3+][SCN-]

The titration must be carried out in acidic solution to prevent precipitation of

iron(III) as the hydrated oxide.

The strong acidic environment required for the Volhard procedure represents a

distinct advantage over other titrimetric methods of halide analysis because such ions

as carbonate, oxalate, and arsenate (which form slightly soluble silver salts in neutral

media but not in acidic media) do not interfere.

Silver chloride is more soluble than silver thiocyanate. As a consequence, in

chloride determinations by the Volhard method, the reaction:

AgCl(s) + SCN- AgSCN(s) + Cl-

occurs to a significant extent near the end of the back-titration of the excess silver ion.

This reaction causes the end point to fade and results in an overconsumption of

thiocyanate ion, which in turn leads to low values for the chloride analysis. This error

can be circumvented by filtering the silver chloride before undertaking the back-

94

titration. Filtration is not required in the determination of other halides because they

all from silver salts that are less soluble than silver thiocyanate. We can also heat the

solution containing the precipitate to collect the particles together, or adding some

organic solvent such as nitrobenzene or chloroform, which covers the AgCl precipitate

in the bottom of conical flask and shields it from the aqueous medium which contains

the excess Ag+.

Another application for Volhard’s method is to determine AgNO3 concentration by

direct titration with thiocyanate using iron(III) ions as indicator.

96

8.1 Introduction

Most metal ions react with electron-pair donors to form coordination compounds

or complexes. The donor species, or ligand, must have at least one pair of unshared

electrons available for bond formation. Water, ammonia, and halide ions are common

inorganic ligands. In fact, most metal ions in aqueous solution actually exist as aquo

complexes. Copper(II) in aqueous solution, for example, is readily complexed by

water molecules to form species such as Cu(H2O)42+. We often simplify such

complexes in chemical equations by writing the metal ion as if it were uncomplexed

Cu2+. Remember that such ions are actually aquo complexes in aqueous solution.

The number of covalent bonds that a cation tends to form with electron donors is

its coordination number. Typical values for coordination number are 2, 4, and 6. The

species formed as a result of coordination can be electrically positive, neutral, or

negative. For example, copper(II), which has a coordination number of 4, forms a

cationic ammine complex, Cu(NH3)42+; a neutral complex with glycine,

Cu(NH2CH2COO)2; and an anionic complex with chloride ion, CuCl42-.

Titrimetric methods based on complex formation, sometimes called

complexometric methods, have been used for more than a century. The truly

remarkable growth in their analytical application, based on a particular class of

coordination compounds called chelates. A chelate is produced when a metal ion

coordinates with two or more donor groups of a single ligand to form a five- or six-

member heterocyclic ring. The copper complex of glycine mentioned is an example.

Here, the copper bonds to both the oxygen of the carboxyl group and the nitrogen of

the amine group.

A ligand that has a single donor group, such as ammonia, is called unidentate

(single-toothed), whereas one such as glycine, which has two groups available for

covalent bonding, is called bidentate. Tridentate, tetradentate, pentadentate, and

hexadentate chelating agents are also known.

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Another important type of complex is formed between metal ions and cyclic

organic compounds, known as macrocycles. These molecules contain nine or more

atoms in the cycle and include at least three heteroatoms, usually oxygen, nitrogen, or

sulfur.

8.2 Complexation Equilibria

Complexation reactions involve a metal ion M reacting with a ligand L to form a

complex ML, as shown in the equation:

M + L ML ……….1)

Where the charges on the ions are omitted so as to be general. Complexation reactions

occur in a stepwise fashion; the reaction in equation 1 is often followed by additional

reactions:

ML + L ML2 …….…2)

ML2 + L ML3 …….…3)

MLn-1 + L MLn …….…4)

Unidentate ligands invariably add in a series of steps, as shown here. With

multidentate ligands, the maximum coordination number of the cation may be satisfied

with only one ligand or a few added ligands. For example, Cu(II), with a maximum

coordination number of 4, can form complexes with ammonia that have formulas

Cu(NH3)2+, Cu(NH3)2

2+, Cu(NH3)32+, and Cu(NH3)4

2+. With the bidentate ligand

glycine (gly), the only complexes that form are Cu(gly)2+ and Cu(gly)22+.

The equilibrium constants for complex formation reactions are generally written as

formation constants. Thus, each of the reactions 1 through 4 is associated with a

stepwise formation constant K1 through K4. For example, K1 = [ML]/[M][L], K2 =

[ML2]/[ML][L], and so on. We can also write the equilibria as the sum of individual

steps. These have overall formation constants designated by the symbol βn. Thus,

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[ML]

M + L ML β1 = ــــــــــــــــــــــ = K1 ………….5)

[M][L]

[ML2]

M + 2L ML2 β2 = ــــــــــــــــــــــ = K1 K2 …...…6)

[M][L]2

[ML3]

M +3L ML3 β3 = ــــــــــــــــــــــ = K1 K2 K3 …….7)

[M][L]3

[MLn]

M + nL MLn βn = ــــــــــــــــــــــ = K1 K2….. Kn ...8)

[M][L]n

Except for the first step, the overall formation constants are products of the stepwise

formation constants for the individual steps leading to the product.

8.3 Titrations with Inorganic Complexing Agents

Complexation reactions have many uses in analytical chemistry, but their classical

application is in complexometric titrations. Here, a metal ion reacts with a suitable

ligand to form a complex, and the equivalence point is determined by an indicator or

an appropriate instrumental method. The formation of soluble inorganic complexes is

not widely used for titrations, but the formation of precipitates, particularly with silver

nitrate as the titrant, is the basis for many important determinations.

The progress of a complexometric titration is generally illustrated by the titration

curve, which is usually a plot of pM = -log [M] as a function of the volume of titrant

added. Most often, in complexometric titration the ligand is the titrant and the metal

ion the analyte, although occasionally the reverse is true. Most simple inorganic

ligands are unidentate, which can lead to low complex stability and indistinct titration

end points. As titrants, multidentate ligands, particularly those having four or six

donor groups, have two advantageous over their unidentate counterparts. First, they

generally react more completely with cations and thus provide sharper end points.

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Second, they ordinarily react with metal ions in a single-step process, whereas

complex formation with unidentate ligands involves two or more intermediate species.

The advantage of a single-step reaction is illustrated by the titration curves shown

in Figure 8-1. Each of the titrations involves a reaction that has an overall equilibrium

constant of 1020. Curve A is derived for a reaction in which a metal ion M having a

coordination number of 4 reacts with a tetradentate ligand D to form the complex of

MD. (We have again omitted the charges on the two reactants for convenience). Curve

B is for the reaction of M with a hypothetical bidentate ligand B to give MB2 in two

steps. The formation constant for the first step is 1012 and for the second 108. Curve C

involves a unidentate ligand A that forms MA4 in four steps with successive formation

constant of 108, 106, 104, and 102. These curves demonstrate that a much sharper end

point is obtained with a reaction that takes place in a single step. For this reason,

multidentate ligands are ordinarily preferred for complexometric titration.

Figure 8-1 Titration Curves for Complexometric Titrations. Titration of 60.0 mL of a solution that is

0.020 M in metal M with (A) a 0.020 M solution of the tetradentate ligand D to give MD as the product;

(B) a 0.040 M solution of the bidentate ligand B to give MB2; and (C) a 0.080 M solution of the

unidentate ligand A to give MA4. The overall formation constant for each product is 1020.

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The most widely used complexometric titration employing a unidentate ligand is

the titration of cyanide with silver nitrate. This method involves the formation of the

soluble Ag(CN)2-. Other titrant are Hg(NO3)2 for analyte Br-, Cl-, SCN-, CN-, thiourea,

and NiSO4 for CN-, and KCN for Cu2+, Hg2+, Ni2+.

8.4 Organic Complexing Agents

Many different organic complexing agents have become important in analytical

chemistry because of their inherent sensitivity and potential selectivity in reacting with

metal ions. Such reagents are particularly useful in precipitating metals, in binding

metals to prevent interferences, in extracting metals from one solvent to another, and

in forming complexes that absorb light for spectrophotometric determinations. The

most useful organic reagents form chelate complexes with metal ions.

Many organic reagents are used to convert metal ions into forms that can be

readily extracted from water into an immiscible organic phase as shown in Table 8-1.

Table 8-1 Organic Reagents for Extracting Metals

Reagent Metal Ions Extracted Solvents

8-Hydroxyquinoline Zn2+, Cu2+, Ni2+,

Al3+

Water CHCl3

Diphenylthiocarbazone Cd2+, Co2+, Cu2+,

Pb2+

Water CHCl3, or CCl4

Acetylacetone Fe3+, Cu2+, Zn2+,

U(VI)

Water CHCl3, CCl4,or

C6H6

Ammonium pyrrolidine

dithiocarbamate Transition metals

Water Methyl isobutyl

ketone

* Ethylenediaminetetraacetic acid (EDTA)

It is the most widely used complexometric titrant. EDTA has the structural

formula:

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The EDTA molecule has six potential sites for bonding a metal ion: the four

carboxylic groups and the two amino groups, each of the latter with an unshared pair

of electrons. Thus, EDTA is a hexadentate ligand.

The dissociation constants for the acidic groups in EDTA are K1 = 1.02 × 10-2, K2

= 2.14 × 10-3, K3 = 6.92 × 10-7, and K4 = 5.50 × 10-11. It is of interest that the first two

constants are of the same order of magnitude, which suggests that the two protons

involved dissociate from opposite ends of the rather long molecule. As a consequence

of their physical separation, the negative charge created by the first dissociation does

not greatly affect the removal of the second proton. The same cannot be said for the

dissociation of the other two protons, however, which are much closer to the

negatively charged carboxylate ions created by the initial dissociations.

The various EDTA species are often abbreviated H4Y, H3Y-, H2Y

2-, HY3-, and Y4-.

* Reagents for EDTA titration

The free acid H4Y and the dihydrate of the sodium salt, Na2H2Y.2H2O, are

commercially available in reagent quality. The former can serve as a primary standard

after it has been dried for several hours at 130oC to 145oC. It is then dissolved in the

minimum amount of base required for complete solution.

* Complexes of EDTA and metal ions

Solutions of EDTA are particularly valuable as titrants because the reagent

combines with metal ion in a 1:1 ratio regardless of the charge on the cation. For

example, the silver and aluminum complexes are formed by the reactions:

Ag+ + Y4- AgY3-

Al3+ + Y4- AlY-

EDTA is a remarkable reagent not only because it forms chelates with all cations

except alkali metals but also because most of these chelates are sufficiently stable for

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titrations. This great stability result from the several complexing sites within the

molecule that give rise to a cage-like structure, in which the cation is surrounded by

and isolated from solvent molecules.

* Effect of other complexing agents on EDTA

Many EDTA titrations are complicated by the tendency on the part of the ion

being titrated to precipitate as a basic oxide or hydroxide at the pH required for

satisfactory titration. In order to keep the metal ion in solution, particularly in the early

stage of the titration, it is necessary to include an auxiliary complexing agent. Thus,

for example, the titration of zinc(II) is ordinary performed in the presence of high

concentrations of ammonia and ammonium chloride. These species buffer the solution

to an acceptable pH. In addition, the ammonia prevents precipitation of zinc hydroxide

by forming ammine complexes. The titration reaction with EDTA is thus best

represented as:

Zn(NH3)42+ + HY3- ZnY2- + 3NH3 + NH4

+

The solution also contains such other zinc/ammonia species as Zn(NH3)32+,

Zn(NH3)22+, and Zn(NH3)

2+. Calculation of pZn in a solution that contains ammonia

must take these species into account. Qualitatively, complexation of a cation by an

auxiliary complexing reagent causes preequivalence pM values to be larger than in a

comparable solution with no such reagent.

* Indicators for EDTA titrations

Nearly 200 organic compounds have been investigated as indicators for metal ions

in EDTA titrations. In general these indicators are organic dyes that form colored

chelate with metal ions in a pM range that is characteristic of the particular cation dye.

The complexes are often intensely colored and are discernible to the eye at

concentrations in the range of 10-6 to 10-7 M.

Eriochrome Black T is a typical metal ion indicator that is used in the titration of

several common cations. Its behavior as a weak acid is described by the equations:

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H2O + H2In- HIn2- + H3O

+ K1 = 5 × 10-7 red blue

H2O + HIn2- In3- + H3O+ K2 = 2.8 × 10-12

blue orange

Note that the acids and their conjugate bases have different colors. Thus, Eriochrome

Black T behaves as an acid-base indicator as well as a metal ion indicator.

The metal complexes of Eriochrome Black T are generally red, as is H2In-. Thus,

for metal ion detection, it is necessary to adjust the pH to 7 or above so that the blue

form of the species, HIn2-, predominates in the absence of a metal ion. Until the

equivalence point in a titration, the indicator complexes the excess metal ion so that

the solution is red. With the first slight excess of EDTA, the solution turns blue as a

consequence of the reaction:

MIn- + HY3- HIn2- + MY2- red blue

Eriochrome Black T forms red complexes with more than two dozen metal ions,

but the formation constant of only a few are appropriate for end point detection.

8.5 The Scope of EDTA Titrations

Complexometric titrations with EDTA have been applied to the determination of

virtually every metal cation, with the exception of the alkali metal ions. Because

EDTA complexes most cations, the reagent might appear at first glance to be totally

lacking in selectivity. In fact, however, considerable control over interferences can be

realized by pH regulation. For example, trivalent cations can usually be titrated

without interference from divalent species by maintaining the solution at a pH of about

1. At this pH, the less stable divalent chelates do not form to any significant extent, but

the trivalent ions are quantitatively complexed.

Similarly, ions such as cadmium and zinc, which form more stable EDTA chelate

than does magnesium, can be determined in the presence of the latter ion by buffering

the mixture to a pH of 7 before titration. Eriochrome Black T serves as an indicator for

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the cadmium or zinc end points without interference from magnesium because the

indicator chelate with magnesium is not formed at this pH.

Finally, interference from a particular cation can sometimes be eliminated by

adding a suitable masking agent, an auxiliary ligand that preferentially forms highly

stable complexes with the potential interfering ion. Thus, cyanide ion is often

employed as a masking agent to permit the titration of magnesium and calcium ions in

the presence of ions such as cadmium, cobalt, copper, nickel, zinc, and palladium. All

of the latter form sufficiently stable cyanide complexes to prevent reaction with EDTA

8.6 The Determination of Water Hardness

Water hardness was defined in terms of the capacity of cations in the water to

replace the sodium or potassium ions in soaps and to form sparingly soluble products

that cause scum in the sink or bathtub. Most multiply charged cations share this

undesirable property. In natural waters, the concentrations of calcium and magnesium

ions generally far exceed those of any other metal ion. Consequently, hardness is now

expressed in terms of the concentration of calcium carbonate that is equivalent to the

total concentration of all the multivalent cations in the sample.

The determination of hardness is a useful analytical test that provides a measure of

the quality of water for household and industrial uses. The test is important to industry

because hard water, on being heated, precipitates calcium carbonate, which clogs

boilers and pipes.

Water hardness is ordinarily determined by EDTA titration after the sample has

been buffered to pH 10. Magnesium, which forms the least stable EDTA complex of

all of the common multivalent cations in typical water samples, is not titrated until

enough reagent has been added to complex all of the other cations in the sample.

Therefore, a magnesium ion indicator, such as Calmagite or Eriochrome Black T, can

serve as indicator in water hardness titrations. Often, a small concentration of the

magnesium-EDTA chelate is incorporated into the buffer or the titrant to ensure

sufficient magnesium ions for satisfactory indicator action.

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9.1 Equilibria in Oxidation-Reduction system

Oxidation-reduction, or redox, processes involve the transfer of electrons from one

reactant to another. Volumetric methods based upon electron transfer are more

numerous and more varied than those for any other reaction type.

Oxidation involves the loss of electrons by a substance and reduction the gain of

electrons. In any oxidation-reduction reaction, the molar ratio between the substance

oxidized and the substance reduced is such that the number of electrons lost by one

species is equal to the number gained by the other. This fact must always be taken into

account when balancing equations for oxidation-reduction reactions.

Oxidizing agents or oxidants possess a strong affinity for electrons and cause other

substances to be oxidized by abstracting electrons from them. In the process, the

oxidizing agent accepts electrons and is thereby reduced. Reducing agents or

reductants have little affinity for electrons and, in fact, readily give up electrons

thereby causing some other species to be reduced. A consequence of this electrons

transfer is the oxidation of the reducing agent.

Separation of an oxidation-reduction reaction into its component parts (that is, into

half-reactions) is a convenient way of indicating clearly the species that gains

electrons and the one that losses them. Thus, the overall reaction:

5Fe2+ + MnO4- + 8H+ 5Fe3+ + Mn2+ + 4H2O

is obtained by combining the half-reaction for the oxidation of iron(II):

5Fe2+ 5Fe3+ + 5e-

with that for the reduction of permanganate:

MnO4- + 5e- + 8H+ Mn2+ + 4H2O

Note that it was necessary to multiply the first half-reaction through by 5 to eliminate

the electrons from the overall equation.

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9.2 Balancing of Redox Equations

The redox equations can be balanced by two methods:

* The change in oxidation numbers method

By this method we follow the rules:

1- Determine the oxidation numbers for all the elements involved in the reaction.

2- From the oxidation numbers determine the oxidizing and reducing agents.

3- Count the difference in the oxidation numbers from the oxidizing and reducing

agents, and according to it chose the numbers for alternate multiplication, in order to

equalize the number of lost and gained electrons.

4- Balance the other ions that did not change.

Example 1

Balance the following equation:

KMnO4 + NH3 KNO3 + MnO2 + KOH + H2O

We determine the oxidation numbers:

KMnO4 + NH3 KNO3 + MnO2 + KOH + H2O

+1 +7 (-2)4 -3 (+1)3 +1 +5 (-2)3 +4 (-2)2 +1 -2 +1 (+1)2 -2

Note the change in the manganese and nitrogen oxidation numbers.

Mn7+ Mn4+ the change (-3)

(alternate multiplication)

N3- N5+ the change (+8)

To maintain the total number of oxidation numbers, we need 8 manganese atoms for

each 3 nitrogen atoms.

8KMnO4 + 3NH3 3KNO3 + 8MnO2 + KOH + H2O

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The oxidation number of potassium did not change, so we balance it and also we

balance the oxygen and hydrogen atoms.

8KMnO4 + 3NH3 3KNO3 + 8MnO2 + 5KOH + 2H2O

* The ion-electron method

To use this method we follow the rules:

1- Divide the reaction to oxidation reaction and reduction reaction, and write it in ionic

form.

2- Balance the number of atoms for each half-reaction.

3- Balance each half-reaction electrically by the addition of a suitable number of

electrons to the more electropositive side.

4 - The number of electrons in each half-reaction must be equal, if not use the

alternate multiplication.

5- Add the two balanced half reactions to each other.

6- The molecular equation can be obtained from the ionic equation by addition of ions

with the reactants and distribute it over the products.

7- We may need to add water molecules or its ions (H+, OH-) to balance the equation.

Example 2

Balance the following equation:

K2Cr2O7 + FeCl2 + HCl CrCl3 + FeCl3 + H2O + KCl

The equation in the ionic form is:

Cr2O72- + Fe2+ + H+ Cr3+ + Fe3+ + H2O

The reduction half-reaction is:

Cr2O72- + H+ Cr3+ + H2O

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To balance the number of atoms we multiply Cr3+ by 2, water by 7, and hydrogen by

14, the equation becomes:

Cr2O72- + 14H+ 2Cr3+ + 7H2O

Now we balance the electrons on each side of the reaction. The number of positive

charges in the reactants is 12 and in the products is 6 only, and to reduce 12 positive

charges to 6 we need to add 6 electrons to the reaction:

Cr2O72- + 14H+ + 6e- 2Cr3+ + 7H2O

The number of atoms and charges in this half reaction is balanced.

The oxidation half-reaction is:

Fe2+ Fe3+

The number of atoms is equal, but the number of charges is not, and to balance it we

add 1 electron to the more electropositive side.

Fe2+ Fe3+ + e-

and to make the number of electrons equal we multiply the last equation by 6, then add

the two equations:

Cr2O72- + 14H+ + 6e- 2Cr3+ + 7H2O

6Fe2+ 6Fe3+ + 6e-

addition ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ

Cr2O72- + 6Fe2+ + 14H+ 2Cr3+ + 6Fe3+ + 7H2O

To obtain the molecular equation, we add the other ions and distribute it over the

products:

K2Cr2O7 + 6FeCl2 +14HCl 2CrCl3 + 6FeCl3 + 7H2O + 2KCl

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9.3 Auxiliary Oxidizing and Reducing Reagents

The analyte in an oxidation-reduction titration must be in a single oxidation state at

the outset. Often, the steps that precede the titration, such as dissolving the sample and

separating interferences, convert the analyte to a mixture of oxidation states. For

example, when a sample containing iron is dissolved, the resulting solution usually

contains a mixture of iron(II) and iron(III) ions. If we choose to use a standard oxidant

to determine iron, we must first treat the sample solution with an auxiliary reducing

agent to convert all of the iron to iron(II). If we plan to titrate with a standard

reductant, however, pretreatment with an auxiliary oxidizing reagent is needed.

To be useful as a preoxidant or a prereductant, a reagent must react quantitatively

with the analyte. In addition, any reagent excess must be easily removable because the

excess reagent usually interferes with the titration by reacting with the standard

solution.

9.4 Auxiliary Reducing Reagents

A number of metals are good reducing agents and have been used for the

preredution of analytes. Included among these are zinc, aluminum, cadmium, lead,

nickel, copper, and silver (in the presence of chloride ion). Sticks or coil of the metal

can be immersed directly in the analyte solution. After reduction is judged complete,

the solid is removed manually and rinsed with water. The analyte solution must be

filtered to remove granular or powdered forms of the metal. An alternative to filtration

is the use of a redactor, such as Jones redactor, which has a diameter of about 2 cm

and holds a 40 to 50 cm column of amalgamated zinc.

Zinc can be amalgamated by allowing its granules to stand briefly in a solution of

mercury(II) chloride, where the following reaction occurs:

2Zn(s) + Hg2+ Zn2+ + Zn(Hg)(s)

Zinc amalgam is nearly as effective for reduction as the pure metal and has the

important virtue of inhibiting the reduction of hydrogen ions by zinc. This side

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reaction needlessly uses up the reducing agent and also contaminates the sample

solution with a large amount of zinc(II) ions. Solutions that are quite acidic can be

passed through a redactor without significant hydrogen formation.

9.5 Auxiliary Oxidizing Reagents

* Sodium Bismuthate

Sodium bismuthate is a powerful oxidizing agent capable, for example, of

converting manganese(II) quantitatively to permanganate ion. This bismuth salt is a

sparingly soluble solid with a formula that is usually written as NaBiO3, although its

exact composition is somewhat uncertain. Oxidations are performed by suspending the

bismuthate in the analyte solution and boiling for brief period. The unused reagent is

then removed by filtration. The half-reaction for the reduction of sodium bismuthate

can be written as:

NaBiO3(s) + 4H+ + 2e- BiO+ + Na+ + 2H2O

* Ammonium Peroxydisulfate

Ammonium peroxydisulfate, (NH4)2S2O8, is also a powerful oxidizing agent. In

acidic solution, it convert chromium(III) to dichromate, cerium(III) to cerium(IV), and

manganese(II) to permanganate. The half-reaction is:

S2O82- + 2e- 2SO4

2-

The oxidations are catalyzed by trace of silver ion. The excess reagent is easily

decomposed by a brief period of boiling:

2S2O82- + 2H2O 4SO4

2- + O2(g) + 4H+

* Sodium Peroxide and Hydrogen Peroxide

Peroxide is a convenient oxidizing agent either as the sodium salt or as a dilute

solution of the acid. The half-reaction for hydrogen peroxide in acidic solution is:

H2O2 + 2H+ + 2e- 2H2O

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After oxidation is complete, the solution is freed of excess reagent by boiling:

2H2O2 2H2O + O2(g)

9.6 Applying Standard Reducing Agents

Standard solutions of most reductants tend to react with atmospheric oxygen. For

this reason, reductants are seldom used for direct titration of oxidizing analytes;

indirect methods are used instead. The two most common reductants are iron(II) and

thiosulfate ions.

* Iron(II) solutions

Solutions of iron(II) are easily prepared from iron(II) ammonium sulfate,

Fe(NH4)2(SO4)2.6H2O (Mohr's salt), or from the closely related iron(II)

ethylenediamine sulfate FeC2H4(NH3)2(SO4)2.4H2O (Oesper's salt). Air oxidation of

iron(II) takes place rapidly in neutral solutions but is inhibited in the presence of acids,

with most stable preparations being about 0.5 M in H2SO4. Such solutions are stable

for no more than one day. Numerous oxidizing agents are conveniently determined by

treatment of the analyte solution with a measured excess of standard iron(II) followed

by immediate titration of the excess with a standard solution of potassium dichromate

or cerium(IV). Just before or just after the analyte is titrated, the volumetric ratio

between the standard oxidant and the iron(II) solution is established by titrating two or

three aliquots of the latter with former.

This procedure has been applied to the determination of organic peroxides;

hydroxylamine; chromium(VI); cerium(IV); molybdenum(VI); nitrate, chlorate, and

perchlorate ions; and numerous other oxidants.

* Sodium Thiosulfate

Thiosulfate ion (S2O32-) is a moderately strong reducing agent that has been widely

used to determine oxidizing agents by an indirect procedure that involves iodine as an

intermediate. With iodine, thiosulfate ion is oxidized quantitatively to tetrathionate ion

(S4O62-) according to half-reaction:

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2S2O32- S4O6

2- + 2e-

The quantitative reaction with iodine is unique. Other oxidants can oxidize the

tetrathionate ion to sulfate ion.

The scheme used to determine oxidizing agents involves adding an unmeasured

excess of potassium iodide to a slightly acidic solution of the analyte. Reduction of the

analyte produces a stoichiometrically equivalent amount of iodine. The liberated

iodine is then titrated with a standard solution of sodium thiosulfate, Na2S2O3, one of

the few reducing agents that is stable toward air oxidation. An example of this

procedure is the determination of sodium hypochlorite in bleaches. The reactions are:

OCl- + 2I- + 2H+ Cl- + I2 + H2O (unmeasured excess KI)

I2 + S2O32- 2I- + S4O6

2-

The quantitative conversion of thiosulfate ion to tetrathionate ion requires a pH

smaller than 7. If strongly acidic solution must be titrated, air oxidation of the excess

iodide must be prevented by blanking the solution with an inert gas, such as carbon

dioxide or nitrogen.

*Detecting end points in iodine-thiosulfate titration

A solution that is about 5×10-6 M in I2 has a discernible color, which corresponds

to less than one drop of a 0.05 M iodine solution in 100 mL. Thus, provided that the

analyte solution is colorless, the disappearance of the iodine color can serve as the

indicator in titrations with sodium thiosulfate.

More commonly, titrations involving iodine are performed with a suspension of

starch as an indicator. The deep blue color that develops in the presence of iodine is

believed to rise from the absorption of iodine into the helical chain of β-amylose, a

macromolecular component of most starches. The closely related α-amylose forms a

red adduct with iodine. This reaction is not easily reversible and is thus undesirable. In

114

commercially available soluble starch, the alpha fraction has been removed to leave

principally β-amylose; indicator solutions are easily prepared from this product.

Aqueous starch suspensions decompose within a few days, primarily because of

bacterial action. The decomposition products tend to interfere with the indicator

properties of the preparation and may also be oxidized by iodine. The rate of

decomposition can be inhibited by preparing and storing the indicator under sterile

conditions and by adding mercury(II) iodide or chloroform as a bacteriostat.

*Standardizing thiosulfate solution

Potassium iodate is an excellent primary standard for thiosulfate solutions. In this

application, weighed amounts of primary-standard grade reagent are dissolved in

water containing an excess of potassium iodide. When this mixture is acidified with a

strong acid, the following reaction occurs instantaneously:

IO3- + 5I- + 6H+ 3I2 + 2H2O

The liberated iodine is then titrated with the thiosulfate solution. The stoichiometry of

the reaction is:

1 mol IO3- = 3 mol I2 = 6 mol S2O3

2-

Other primary standards for sodium thiosulfate are potassium dichromate, potassium

bromate, potassium hydrogen iodate, potassium hexacyanoferrate(III), and metallic

copper. All these compounds liberate stoichiometric amounts of iodine when treated

with excess potassium iodide.

9.7 Applying Standard Oxidizing Agents

The choice among oxidizing agents depends on the strength of the analyte as a

reducing agent, the rate of reaction between oxidant and analyte, the stability of the

standard oxidant solutions, the cost, and the availability of a satisfactory indicator.

* The strong oxidants–potassium permanganate and cerium(IV)

Solutions of permanganate ion and cerium(IV) ion are strong oxidizing reagents

whose applications closely parallel one another. Half-reactions for the two are:

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MnO4- + 8H+ + 5e- Mn2+ + 4H2O Eo = 1.51 V

Ce4+ + e- Ce3+ Eo = 1.44 V (1 M H2SO4)

The formal potential shown for the reaction of cerium(IV) is for solutions that are 1 M

in sulfuric acid. In 1 M perchloric acid and 1 M nitric acid, the potentials are 1.70 V

and 1.61 V, respectively. Solutions of cerium(IV) in the latter two acids are not very

stable and thus find limited application.

The half-reaction shown for permanganate ion occurs only in solutions that are

0.1 M or greater in strong acid. In less acidic media, the product may be Mn(III),

Mn(IV), or Mn(VI), depending on conditions.

Despite some advantages of cerium solutions, like stability, over permanganate

solutions, the latter are more widely used. One reason is the color of permanganate

solutions, which is intense enough to serve as an indicator in titrations. A second

reason for the popularity of permanganate solutions is their modest cost.

*Detecting the end points

A useful property of a potassium permanganate solution is its intense purple color,

which is sufficient to serve as an indicator for most titrations. If you add as little as

0.01 to 0.02 mL of a 0.02 M solution of permanganate to 100 mL of water, you can

perceive the purple color of the resulting solution. If the solution is very dilute,

diphenylamine sulfonic acid or the 1,10-phenanthroline complex of iron(II) provides a

sharper end point.

The permanganate end point is not permanent because excess permanganate ions

react slowly with the relatively larger concentration of manganese(II) ions present at

the end point, according to the reaction:

2MnO4- + 3Mn2+ + 2H2O 5MnO2(s) + 4H+

The equilibrium constant for this reaction is about 1047, which indicates that the

equilibrium concentration of permanganate ion is incredibly small even in highly

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acidic media. Fortunately, the rate at which this equilibrium is approached is so slow

that the end point fades only gradually over a period of perhaps 30 seconds.

Solutions of cerium(IV) are yellow-orange, but the color is not intense enough to

act as an indicator in titrations. Several oxidation-reduction indicators are available for

titrations with standard solutions of cerium(IV). The most widely used of these is the

iron(II) complex of 1,10-phenanthroline or one of its substituted derivatives.

*The preparation and stability of standard solutions

Aqueous solutions of permanganate are not entirely stable because of water

oxidation:

4MnO4- + 2H2O 4MnO2(s) + 3O2(g) + 4OH-

Although the equilibrium constant for this reaction indicates that the products are

favored, permanganate solutions, when properly prepared, are reasonably stable

because the decomposition reaction is slow. It is catalyzed by light, heat, acids, bases,

manganese(II), and manganese dioxide.

Moderately stable solutions of permanganate ion can be prepared if the effects of

these catalyst, particularly manganese dioxide, are minimized. Manganese dioxide is a

contaminant in even the best grade of solid potassium permanganate. Removal of

manganese dioxide by filtration before standardization markedly improves the stability

of standard permanganate solutions. Before filtration, the reagent solution is allowed

to stand for about 24 hours or is heated for a brief period to hasten oxidation of the

organic species generally present in small amounts in distilled and deionized water.

Paper cannot be used for filtration because permanganate ion reacts with it to form

additional manganese dioxide.

Standardized permanganate solutions should be stored in the dark. Filtration and

restandardization are required if any solid is detected in the solution or on the walls of

the storage bottle. In any event, restandardization every 1 or 2 weeks is a good

precautionary measure.

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The most widely used compounds for the preparation of solution of cerium(IV) are

cerium(IV) ammonium nitrate Ce(NO3)4.2NH4NO3, cerium(IV) ammonium sulfate

Ce(SO4)2.2(NH4)2SO4.2H2O, cerium(IV) hydroxide Ce(OH)4, Ce(IV) hydrogen sulfate

Ce(HSO4)4. Primary-standard cerium ammonium nitrate is available commercially and

can be used to prepare standard solutions of the cation directly by weigh. More

commonly, less expensive reagent-grade cerium(IV) ammonium nitrate or ceric

hydroxide is used to prepare solutions that are subsequently standardized. In either

case, the reagent is dissolved in a solution that is at least 0.1 M in sulfuric acid to

prevent the precipitation of basic salts.

Sulfuric acid solutions of cerium(IV) are remarkably stable and can be stored for

months or heated at 100oC for prolonged periods without a change in concentration.

*Standardizing permanganate and cerium(IV) solutions

Sodium oxalate is a widely used primary standard. In acidic solutions, the oxalate

ion is converted to the undissociated acid. Thus, its reaction with permanganate can be

described by:

2MnO4- + 5H2C2O4 + 6H+ 2Mn2+ + 10CO2(g) + 8H2O

The reaction between permanganate ion and oxalic acid is complex and proceeds

slowly even at elevated temperature unless manganese(II) is present as a catalyst.

Thus, when the first few milliliters of standard permanganate are added to a hot

solution of oxalic acid, several seconds are required before the color of the

permanganate ion disappears. As the concentration of manganese(II) builds up,

however, the reaction proceeds more and more rapidly as a result of autocatalysis.

Sodium oxalate is also widely used to standardize Ce(IV) solutions. The reaction

between Ce4+ and H2C2O4 is:

2Ce4+ + H2C2O4 2Ce3+ + 2CO2(g) + 2H+

Cerium(IV) standardization against sodium oxalate are usually performed at 50oC in a

hydrochloric acid solution containing iodine monochloride as a catalyst.

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* Potassium dichromate

In its analytical applications, dichromate ion is reduced to green chromium(III) ion:

Cr2O72- + 14H+ + 6e- 2Cr3+ + 7H2O Eo = 1.33 V

Dichromate titrations are generally carried out in solutions that are about 1 M in

hydrochloric or sulfuric acid. In these media, the formal potential for the half-reaction

is 1.0 to 1.1 V.

Potassium dichromate solutions are indefinitely stable, can be boiled without

decomposition, and do not react with hydrochloric acid. Moreover, primary standard

reagent is available commercially and at modest coast.

*Preparing dichromate Solution

For most purposes, reagent-grade potassium dichromate is sufficiently pure to

permit the direct preparation of standard solutions; the solid is simply dried at 150oC

to 2000C before being weighed.

The orange color of a dichromate solution is not intense enough for use in end

point detection. Diphenylamine sulfonic acid is an excellent indicator for titrations

with this reagent, however. The oxidized form of the indicator is violet, and its

reduced form is essentially colorless; thus, the color change observed in a direct

titration is from the green of chromium(III) to violet.

*Applying potassium dichromate solutions

The principal use of dichromate is the volumetric titration of iron(II) based on the

reaction:

Cr2O72- + 6Fe2+ 14H+ 2Cr3+ + 6Fe3+ + 7H2O

Often, this titration is performed in the presence of moderate concentration of

hydrochloric acid.

The reaction of dichromate with iron(II) has been widely used for the indirect

determination of a variety of oxidizing agents. In these applications, a measured

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excess of an iron(II) solution is added to an acidic solution of the analyte. The excess

iron(II) is then back-titrated with standard potassium dichromate. Standardization of

the iron(II) solution by titration with the dichromate is performed concurrently with

the determination because solution of iron(II) tend to be air oxidized. This method has

been applied to the determination of nitrate, chlorite, permanganate, and dichromate

ions as well as organic peroxides and several other oxidizing agents.

* Iodine

Iodine is a weak oxidizing agent used primarily for the determination of strong

reductants. The most accurate description of the half-reaction for iodine in these

applications is:

I3- + 2e- 3I- Eo = 0.536 V

where I3- is the triiodide ion.

Standard iodine solutions have relatively limited application compared with the

other oxidants we have described because of their significantly smaller electrode

potential. Occasionally, however, this low potential is advantageous because it imparts

a degree of selectivity that makes the determination of strong reducing agents in the

presence of weak ones. Iodine solution lack stability, however, and must be

restandardized regularly.

*Standardizing and applying iodine solution

Iodine solutions can be standardized against anhydrous sodium thiosulfate or

barium thiosulfate monohydrate, both of which are available commercially. Often,

solutions of iodine are standardized against solutions of sodium thiosulfate that have

in turn been standardized against potassium iodate or potassium dichromate.

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10.1 Introduction

Measurement based on light and other forms of electromagnetic radiation are

widely used throughout analytical chemistry. The interaction of radiation and matter

are the subject of the science called spectroscopy. Spectroscopic analytical methods

are based on measuring the amount of radiation produced or absorbed by molecular or

atomic species of interest.

We can classify spectroscopic methods according to the region of the electromagnetic

spectrum involved in the measurement. The regions of the spectrum that have been

used include γ-ray, X-ray, ultraviolet (UV), visible, infrared (IR), microwave, and

radio-frequency (RF). Indeed, current usage extends the meaning of spectroscopy

further to include techniques that do not even involve electromagnetic radiation, such

as acoustic, mass, and electron spectroscopy.

10.2 Electromagnetic Radiation

Electromagnetic radiation, or light, is a form of energy whose behavior is

described by the properties of both waves and particles. The optical properties of

electromagnetic radiation, such as diffraction, are explained best by describing light as

a wave. Many of the interactions between electromagnetic radiation and matter, such

as absorption and emission, however, are better described by treating light as a

particle, or photon. Nevertheless, the dual models of wave and particle behavior

provide a useful description for electromagnetic radiation.

10.3 Wave Properties

In dealing with phenomena such as reflection, refraction, interference, and

diffraction, electromagnetic radiation is conveniently modeled as wave consisting of

perpendicularly oscillating electric and magnetic fields. The electric field for a single-

frequency wave oscillates sinusoidally in space and time, as shown in Figure 10-1.

The electric field is represented as a vector whose length is proportional to the field

strength.

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Figure 10-1 Radiation of Wavelength λ and Amplitude A. The arrows represent the electric

vector of the radiation.

The x-axis in this plot is either time as the radiation passes a fixed point in space or

distance at a fixed time. Note that the direction in which the field oscillates is

perpendicular to the direction in which the radiation propagates.

10.4 Wave Characteristics

In Figure 10-1, the amplitude of the sine wave is shown, and the wavelength is

defined. The time in seconds required for the passage of successive maxima or minima

through a fixed point in space is called the period, p, of the radiation. The frequency,

υ, is the number of oscillations of the electric field vector per unit time and is equal to

1/p.

The frequency of a light wave, or any wave of electromagnetic radiation, is

determined by the source that emits it and remains constant regardless of the medium

traversed. In contrast, the velocity, v, of the wave front through a medium depends on

both the medium and the frequency. The wavelength, λ, is the linear distance between

successive maxima or minima of a wave, as shown in Figure 10-1. Multiplication of

the frequency (in waves per unit time) by the wavelength (in distance per wave) gives

the velocity of the wave, in distance per unit time (cm s-1 or m s-1), as shown in the

equation:

v = υ λ ………1)

Table 10-1 gives the wavelength units for several spectral regions.

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Table 10-1 Wavelength Units for Various Regions

Region Unit Definition

X-ray Angstrom, Ao 10-10 m

Ultraviolet/visible Nanometer, nm 10-9 m

Infrared Micrometer, µm 10-6 m

10.5 The Speed of Light

In a vacuum, light travels at its maximum velocity. This velocity, which is given

the special symbol c, is 2.99792 × 108 m s-1. The velocity of light in air is only about

0.03% less than its velocity in a vacuum. Thus, for a vacuum or for air, Equation 1

conveniently gives the velocity of light.

c = υ λ ....…….2)

= 3.00 × 108 m s-1 = 3.00 × 1010 cm s-1

In the medium containing matter, light travels, with a velocity less than c because

of interaction between the electromagnetic field and electrons in the atoms or

molecules of the medium. Since the frequency of the radiation is constant, the

wavelength must decrease as the light passes from a vacuum to a medium containing

matter.

The wavenumber ῡ is another way to describe electromagnetic radiation. It is

defined as the number of waves per centimeter and is equal to 1/λ. By definition, ῡ has

the units of cm-1.

10.6 Radiant Power and Intensity

The radiant power p in watts (W) is the energy of a beam that reaches a given area

per unit time. The intensity is the radiant power-per-unit solid angle (Solid angle is the

three dimensional spread at the vertex of a cone measured as the area intercepted by

the cone on a unit sphere whose center is at the vertex). Both quantities are

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proportional to the square of the amplitude of the electric field. Although it is not

strictly correct, radiation power and intensity are frequently used interchangeably.

10.7 The Particle Nature of Light: Photons

In many radiation/matter interactions, it is most useful to consider light as

consisting of photons or quanta. We can relate the energy of a photon to its

wavelength, frequency, and wavenumber by:

hc

E = hυ = ــــــــــــ = hcῡ ……….3)

λ

where h is Planck's constant (6.63 × 10-34 J s). Note that the wavenumber and

frequency, in contrast to the wavelength, are directly proportional to the photon

energy. Wavelength is inversely proportional to energy. The radiant power of a beam

of radiation is directly proportional to the number of photon per second.

Example 1

Calculate the wavenumber of a beam of infrared radiation with a wavelength of 5.00

µm.

1

ῡ = 2000 = ـــــــــــــــــــــــــــــــــــــــــــــ cm-1

5.00 µm × 10-4 cm/µm

Example 2

Calculate the energy in joules of one photon of the radiation, if the wavenumber is

2000 cm-1.

E = hcῡ

= 6.63 × 10-34 J.s × 3.00 × 1010 cm/s × 2000 cm-1

= 3.98 × 10-20 J

10.8 Interaction of Radiation and Matter

The most interesting types of interactions in spectroscopy involve transitions

between different energy levels of chemical species. Other types of interactions, such

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as reflection, refraction, elastic scattering, interference, and diffraction, are often

related to the bulk properties of materials rather than to energy levels of specific

molecules or atoms. Although these bulk interactions are also of interest in

spectroscopy, we limit our discussion here to those interactions that involve energy-

level transitions. The specific types of interactions that we observe depend strongly on

the energy of the radiation used and the mode of direction.

10.9 The Electromagnetic Spectrum

The electromagnetic spectrum covers an enormous range of energies (frequencies)

and thus wavelengths (Table 10-2). Useful frequencies vary from ˃1019 Hz (γ-ray) to

103 Hz (radio waves). An X-ray photon (υ ≈ 3 × 1018 Hz, λ ≈ 10-10 m), for example, is

approximately 10,000 times as energy as a photon emitted by an ordinary light bulb (υ

≈ 3 × 1014 Hz, λ ≈ 10-6 m) and 1015 times as energetic as a radio-frequency photon (υ ≈

3 × 103 Hz, λ ≈ 105m).

The visible portion, to which our eyes respond, is only a minute portion of the

entire spectrum. Such different types of radiation as gamma (γ) rays or radio waves

differ from visible light in the energy (frequency) of their photons.

Table 10-2 Regions of the UV, Visible, and IR Spectrum

Region Wavelength Range

UV 180-380 nm

Visible 380-780 nm

Near-IR 0.78-2.5 µm

Mid-IR 2.5-50 µm

Figure10-2 shows the regions of the electromagnetic spectrum that are used for

spectroscopic analysis. Also shown are the types of atomic and molecular transition

that result from the interactions of the radiation with sample. Note that the low-energy

radiation used in nuclear magnetic resonance (NMR) and electron spin resonance

(ESR) spectroscopy causes subtle changes, such as changes in spin; the high-energy

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radiation used in γ-ray spectroscopy can produce much more dramatic effects, such as

nuclear configuration changes.

Note that spectrochemical methods that use not only visible but also ultraviolet

and infrared radiation are often called optical methods in spite of the fact that the

human eye is sensitive to neither of the latter two types of radiation.

10.10 Spectroscopic Measurements

Spectroscopists use the interactions of radiation with matter to obtain information

about a sample. Several of the chemical elements were discovered by spectroscopy.

The sample is usually stimulated in some way by applying energy in the form of heat,

electrical energy, light, particles or a chemical reaction. Prior to applying the stimulus,

the analyte is predominantly in its lowest energy state, or ground state. The stimulus

then causes some of the analyte species to undergo a transition to a higher energy or

excited state. We acquire information about the analyte by measuring the

electromagnetic radiation emitted as it returns to the ground state or by measuring the

amount of electromagnetic radiation absorbed as a result of excitation.

Figure 10-2 The Regions of the Electromagnetic Spectrum. Interaction of an analyte with electromagnetic

radiation can result in the types of changes shown.

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Figure 10-3 illustrates the processes involved in emission and chemiluminescence

spectroscopy. Here, the analyte is stimulated by heat or electrical energy or by a

chemical reaction. Emission spectroscopy involves methods in which the stimulus is

heated or electrical energy, while chemiluminescence spectroscopy refers to

excitation of the analyte by a chemical reaction. In both cases, measurement of the

radiant power emitted as the analyte returns to the ground state can give information

about its identity and concentration. The results of such a measurement are often

expressed graphically by a spectrum, which is a plot of the emitted radiation as a

function of frequency or wavelength.

Figure 10-3 Emission or Chemiluminescence Processes. In (a), the sample is excited by the application of thermal,

electrical, or chemical energy. These processes do not involve radiant energy and are hence called nonradiative

processes. In the energy-level diagram (b), the dashed lines with upward pointing arrows symbolize these

nonradiative excitation processes, while the solid lines with downward pointing arrows indicate that the analyte

losses its energy by emission of a photon. In (c), the resulting spectrum is shown as a measurement of the radiant

power emitted PE as a function of wavelength, λ.

When the sample is stimulated by application of an external electromagnetic

radiation source, several processes are possible. For example, the radiation can be

scattered or reflected. What is important to us is that some of the incident radiation can

128

be absorbed and thus promote some of the analyte species to an excited state, as

shown in Figure 10-4.

Figure 10-4 Absorption Methods. In (a) radiation of incident radiant power P0 can be absorbed by the analyte,

resulting in a transmitted beam of lower radiant power P. For absorption to occur, the energy of the incident

beam must correspond to one of the energy differences shown in (b).The resulting absorption spectrum is shown

in (c).

In absorption spectroscopy, we measure the amount of light absorbed as a function

of wavelength. In photoluminescence spectroscopy (Figure 10-5), the emission of

photon is measured after absorption. The most important forms of photoluminescence

for analytical chemistry purposes are fluorescence and phosphorescence

spectroscopy.

Figure 10-5 Photoluminescence Methods (Fluorescence and Phosphorescence). They result from absorption of

electromagnetic radiation and then dissipation of the energy by emission of radiation (a). In (b), the absorption

can cause excitation of the analyte to state 1 or state 2. Once excited, the excess energy can be lost by emission of a

photon (luminescence, shown as solid line) or by nonradiative processes (dashed lines). The emission occurs over

all angles, and the wavelengths emitted (c) correspond to energy differences between levels. The major distinction

between fluorescence and phosphorescence is the time scale of emission, with fluorescence being prompt and

phosphorescence being delayed.

129

10.11 The Absorption of Radiation

The absorption law, also known as the Beer-Lambert law or just Beer's law, tells

us quantitatively how the amount of attenuation depends on the concentration of the

absorbing molecules and the path length over which absorption occurs. As light

traverses a medium containing an absorbing analyte, decreases in intensity occur as

the analyte becomes excited. For an analyte solution of a given concentration, the

longer the length of the medium through which the light passes (path length of light),

the more absorbers are in the path, and the greater the attenuation. Also, for a given

path length of the light, the higher the concentration of absorbers, the stronger the

attenuation.

Figure 10-6 depicts the attenuation of a parallel beam of monochromatic radiation

as it passes through an absorbing solution of thickness b centimeters and concentration

c moles per liter. Because of interaction between the photons and absorbing particles,

the radiant power of the beam decreases from Po to P. The transmittance T of the

solution is the fraction of incident radiation transmitted by the solution. Transmittance

is often expressed as a percentage called the percent transmittance.

T = P / Po …………..4)

The absorbance A of a solution is related to the transmittance in a logarithmic

manner as shown in Equation 5. Notice that as the absorbance of a solution increases,

the transmittance decreases.

P0

A = - log T = log 5.……… ـــــــــــ)

P

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Figure 10-6 Attenuation of a Beam of Radiation by an Absorbing Solution. The larger arrow on the incident beam

signifies a higher radiant power than is transmitted by the solution. The path length of the absorbing solution is b,

and the concentration is c.

10.12 Measuring Transmittance and Absorbance

Ordinarily, transmittance and absorbance, as defined by equations 4 and 5, cannot

be measured because the solution to be studied must be held in some sort of container

(cell or cuvette). Reflection and scattering losses can occur at the cell walls. These

losses can be substantial. For example, about 8.5% of the beam of yellow light is lost

by reflection when it passes through a glass cell. Light can also be scattered in all

directions from the surface of large molecules or particles (such as dust) in the solvent

and this scattering can cause further attenuation of the beam as it passes through the

solution.

The compensate for these effects, the power of the beam transmitted through a cell

containing the analyte solution is compared with one that traverses either an identical

cell containing only the solvent or a reagent blank. An experimental absorbance that

closely approximates the true absorbance for the solution is thus obtained; that is:

P0 Psolvent

A = log ـــــــــــ ≈ log 6……… ـــــــــــــــــــــــ)

P Psolution

The term P0 and P will henceforth refer to the power of a beam that has passed

through cells containing the blank (solvent) and the analyte, respectively.

131

10.13 Beer's Law

According to Beer's law, absorbance is directly proportional to the concentration of

the absorbing species c and to the path length b of the absorbing medium, as expressed

by Equation 7.

A = log (P0/P) = abc ………….7)

Here, a is a proportionality constant called the absorptivity. Because absorbance is a

unitless quantity, the absorptivity must have units that cancel the units of b and c. If,

for example, c has the units of g L-1 and b has the units of cm, absorptivity has the

units of L g-1 cm-1.

When we express the concentration in Equation 7 in moles per liter and b in

centimeters, the proportionality constant is called the molar absorptivity and is given

the special symbol, ε. Thus:

A = εbc

where ε has the units of L mol-1 cm-1.

10.14 Absorption Spectra

An absorption spectrum is a plot of absorbance versus wavelength. Absorbance

could also be plotted against wavenumber or frequency. Most modern scanning

spectrophotometers produce such an absorption spectrum directly.

10.15 Atomic Absorption

When a beam of polychromic ultraviolet or visible radiation passes medium

containing gaseous atoms, only a few frequencies are attenuated by absorption. When

recorded on a very high resolution spectrometer, the spectrum consists of a number of

very narrow absorption lines.

10.16 Emission Spectra

Atoms, ions, and molecules can be excited to one or more higher energy levels by

any of several processes, including bombardment with electron or other elementary

132

particles; exposure to a high-temperature plasma, flame or electric arc; or exposure to

a source of electromagnetic radiation.

Radiation from a source is conveniently characterized by means of an emission

spectrum, which usually takes the form of a plot of the relative power of the emitted

radiation as a function of wavelength or frequency.

Reference

Douglas A. Skoog, Donald M. West, F. Holler, Stanley R. Crouch, Fundamentals of

Analytical Chemistry, Eighth Edition, Brooks/Cole-Thomson Learning. U. S. A. 2004.

1.1 analytical chemistry

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